# Tutor Hunt Questions

“The height h in feet of a model rocket above the ground t seconds after lift-off is given by h(t) =t2+ 100t, for 0 ≤ t ≤ 20.

When does the rocket reach its maximum height above the ground? What is its maximum height?

When does the rocket reach its maximum height above the ground? What is its maximum height?

6 years ago

Maths Question asked by Mrs

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The rocket keeps going up for 20 secs so put t = 20

I am not sure if you have written the equation correctly.

Differentiate the equation and put it equal to zero to find the time t for the maximum height. To find the maximum height, put that value of t into the original equation.

Hello, I can answer this question but would need to clarify something. There appears to be a formatting problem in the question. There is a symbol before the "t2" which appears an an empty square in my browser. Is it supposed to read "h(t) = -t^2 + 100t"? (t^2 means t squared - the ^ is used to indicate a superscri pt in typed mathematics).

the symbol before the `t` appears as a box, what is it meant to be showing? and does `t2` mean t squared? I think the formatting on my Chrome is incorrect

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## 28 Answers

Think about what the gradient of the curve will be at the maximum height. Use differentiation and solve for t. This will give you the time when it happens. Use this to find the height.

Differentiate and equate to zero, rearrange for t to find t which gives max height. Plug same value of t back into original equation to find greatest height.

Simply differentiate the equation and let it equal to 0.

From here you can find the time at max height (when the gradient of the curve = 0).

Input this identified t into the original equation to find the max height.

I can`t answer the equation given to me as I believe the coefficient before the t^2 is missing

From here you can find the time at max height (when the gradient of the curve = 0).

Input this identified t into the original equation to find the max height.

I can`t answer the equation given to me as I believe the coefficient before the t^2 is missing

I think this questions seems hard because of the whole rocket thing and t instead of x. Let`s rephrase it:

a) Solve the equation x^2 + 100x = 0 to find where the graph of y = x^2 + 100x crosses the x axis

b) Sketch the graph of y= x^2 + 100x using the fact that a quadratic is U shaped and goes through the numbers you found in part (a) (ie, 0 and -100)

c) on your sketch, get rid of (scribble out) the bits where x is negative or bigger than 20. We only want to look at the graph from x=0 to x=20

d) look at the sketch - can you see that it is highest when x = 20? Find out how high by subbing x=20 into y = x^2 + 100x to get the y coordinate

They made this look a lot harder by using t instead of x and also talking about a rocket instead of saying whats the highest point on the graph

a) Solve the equation x^2 + 100x = 0 to find where the graph of y = x^2 + 100x crosses the x axis

b) Sketch the graph of y= x^2 + 100x using the fact that a quadratic is U shaped and goes through the numbers you found in part (a) (ie, 0 and -100)

c) on your sketch, get rid of (scribble out) the bits where x is negative or bigger than 20. We only want to look at the graph from x=0 to x=20

d) look at the sketch - can you see that it is highest when x = 20? Find out how high by subbing x=20 into y = x^2 + 100x to get the y coordinate

They made this look a lot harder by using t instead of x and also talking about a rocket instead of saying whats the highest point on the graph

The number in front of the t^2 is missing, therefore I am unable to give you a numerical answer. However, the method that you should use is to devide the range of t by 2 and it will give you 10 since (20-0)/2=10. Now you just need to insert 10 instead of t in your equation which will give you the maximum height in feet (if the question is given in feet). I hope that answered your question.

The formatting of the equation in you question appears corrupted, but is an equation of motion of the form d = u.t + 1/2.a.t^2 (where d is the distance, and in this case height, u is the initial speed, t is the time elapsed, and a is the acceleration which in this case is due to gravity, i.e. 9.8m/s/s). I rather think the corrected form would be d = 100 x t + 1/2 x 9.8 x t^2.

Although the equation comes from physics, the real point of the question is about turning points for a curve, which is mathematics.

The simplest, but more time consuming solution, is to tabulate values and draw a graph. However, if you have done some calculus you should be able to differentiate the equation to determine the gradient of the curve at any time t. As the maximum height will be where the differential of height with respect to time is zero you can use this to solve for when this occurs. (Note: You will need to be careful of the sign convention you use, i.e. consider which direct the distance and acceleration increases in a apply a sign accordingly.)

Although the equation comes from physics, the real point of the question is about turning points for a curve, which is mathematics.

The simplest, but more time consuming solution, is to tabulate values and draw a graph. However, if you have done some calculus you should be able to differentiate the equation to determine the gradient of the curve at any time t. As the maximum height will be where the differential of height with respect to time is zero you can use this to solve for when this occurs. (Note: You will need to be careful of the sign convention you use, i.e. consider which direct the distance and acceleration increases in a apply a sign accordingly.)

You probably mean h(t) = -5(t^2) + 100t

To find the maxima (or minima) you need o find where the gradient of h(t) is zero. To do this, you need to differentiate h(t), see below:

dh / dt = -10t + 100 = 100 - 10t = 10(10 - t)

There is a turning point (a maxima or minima) when the differential is zero (i.e. when the gradient is zero). i.e. when dh/dt = 0

t = 10 seconds

I`m going to assume this is a maxima (if we needed to prove it was a maxima rather than a minima, we`d need to differentiate again, to see if it gave us a positive or negative result, but that is not part of the question)

To find the maxima (or minima) you need o find where the gradient of h(t) is zero. To do this, you need to differentiate h(t), see below:

dh / dt = -10t + 100 = 100 - 10t = 10(10 - t)

There is a turning point (a maxima or minima) when the differential is zero (i.e. when the gradient is zero). i.e. when dh/dt = 0

t = 10 seconds

I`m going to assume this is a maxima (if we needed to prove it was a maxima rather than a minima, we`d need to differentiate again, to see if it gave us a positive or negative result, but that is not part of the question)

its a parabola opening downwards.

The maximum point we need to find is the vertex,where x coordinate represents the time it reaches the maximum height and y coordinate represents the the maximum height it reaches.

x = -b/2a (axis symmetry formula)

x = -100/-2

= 50 seconds

y = -(50)^2 + 100(50)

= -2500 + 5000

= 2500 feet

The maximum point we need to find is the vertex,where x coordinate represents the time it reaches the maximum height and y coordinate represents the the maximum height it reaches.

x = -b/2a (axis symmetry formula)

x = -100/-2

= 50 seconds

y = -(50)^2 + 100(50)

= -2500 + 5000

= 2500 feet

I cannot see all of the equation, so, for now, I will assume that the equation given is:

h(t)= -t^2 + 100t, 0 $leq$ t $leq$ 20

If the coefficients are different the principles are still the same, so don`t sweat.

A good place to start when presented with a quadratic equation is to sketch its corresponding graph. To sketch the graph:

-solve the equation for h=0 to see where the graph crosses the x-axis

-set t=0 to see where the graph crosses the y-axis

-check: is the coefficient of the t^2 term positive or negative?

-if positive our graph shape is a smiley face

-if negative - sad face (which is what we are working with here)

We observe our graph. We notice that the highest point of the graph is at the turning point and that the further we move from this point along the x-axis the smaller the value our graph has against the y-axis. Hence, either the value of t we are looking for is at the turning point or it is the closest value within our range to the turning point. (You may wish to re-read the last two sentences and look at your graph to convince yourself that this is true). (continued)

h(t)= -t^2 + 100t, 0 $leq$ t $leq$ 20

If the coefficients are different the principles are still the same, so don`t sweat.

A good place to start when presented with a quadratic equation is to sketch its corresponding graph. To sketch the graph:

-solve the equation for h=0 to see where the graph crosses the x-axis

-set t=0 to see where the graph crosses the y-axis

-check: is the coefficient of the t^2 term positive or negative?

-if positive our graph shape is a smiley face

-if negative - sad face (which is what we are working with here)

We observe our graph. We notice that the highest point of the graph is at the turning point and that the further we move from this point along the x-axis the smaller the value our graph has against the y-axis. Hence, either the value of t we are looking for is at the turning point or it is the closest value within our range to the turning point. (You may wish to re-read the last two sentences and look at your graph to convince yourself that this is true). (continued)

(continued) So, we start by finding the value of t at the turning point. To do this, we take the derivative of h with respect to t i.e.

t = -t^2 + 100t

dh/dt= -2t + 100

By setting the derivative to 0, we find the value of t at the turning point. So, if

dh/dt = 0 = -2t + 100, then 2t = 100, so t = 50.

We note that 50 > 20 and thus out of our range. We also note that because of our deduction earlier as our extreme value of t=20 is closest to the value of t at the turning point, this must be where our maximum occurs.

Substitute t=20 into our original equation for h:

h= -20^2 + 100*20 = 1600

As a final check, use the other extreme value in our range (t=0) to see if we made a mistake.

h=0 when t=0.

Thus our rocket reaches its max. height when t=20 and h=1600.

Suppose instead that h = -5^2 + 100t:

-similar graph

-dh/dt= -10t + 100 implies t=10 which is in range of values

-sub. t=10 into h to find our max. height.

If h(t) = 5t^2 - 100t why would the max value of h not occur at t=10?

t = -t^2 + 100t

dh/dt= -2t + 100

By setting the derivative to 0, we find the value of t at the turning point. So, if

dh/dt = 0 = -2t + 100, then 2t = 100, so t = 50.

We note that 50 > 20 and thus out of our range. We also note that because of our deduction earlier as our extreme value of t=20 is closest to the value of t at the turning point, this must be where our maximum occurs.

Substitute t=20 into our original equation for h:

h= -20^2 + 100*20 = 1600

As a final check, use the other extreme value in our range (t=0) to see if we made a mistake.

h=0 when t=0.

Thus our rocket reaches its max. height when t=20 and h=1600.

Suppose instead that h = -5^2 + 100t:

-similar graph

-dh/dt= -10t + 100 implies t=10 which is in range of values

-sub. t=10 into h to find our max. height.

If h(t) = 5t^2 - 100t why would the max value of h not occur at t=10?

(Sorry for lack of formatting there appears to be no way to enter a return character, put a return at each "CR")CRCR

The question hasn`t shown all characters so I guess the full equation was;CRCR

h(t) = -0.5at^2+ 100t, for 0 < t < 20. CRCR

That is a standard equation from physics. a is acceleration from gravity. The minus sign recognises that the initial velocity is upwards and gravity pulls it down.CRCR

Early in the flight (low values of t) 100t dominates and the rocket ascends. Late in the flight (high values of t) -0.5at^2 dominates and the rocket descends.CRCR

The "correct" solution is to differentiate height to get velocity;CRCR

V(t) = d h(t)/dt = -at + 100CRCR

At the highest point the rocket changes direction so v(t)=0. Assuming that a=10 m/s^2 (gravity on Earth);CRCR

0 = 100 - at CR

t = 100 / 10 = 10 s CRCR

maximum height is; CRCR

h(t) = 100t - 0.5at^2 CR

= 1000 - 0.5 x 10 x 100 CR

= 500m CRCR

The "cheat" way to do it is to notice that the equation is specified for the time 0 < t < 20. If the rocket flight is symmetrical (true for unpowered flight in a vacuum) it will ascend for the same time that it descends and the maximum height is at the mid point t=10s. The latter is not the textbook solution but allows a quick check on your answer.

The question hasn`t shown all characters so I guess the full equation was;CRCR

h(t) = -0.5at^2+ 100t, for 0 < t < 20. CRCR

That is a standard equation from physics. a is acceleration from gravity. The minus sign recognises that the initial velocity is upwards and gravity pulls it down.CRCR

Early in the flight (low values of t) 100t dominates and the rocket ascends. Late in the flight (high values of t) -0.5at^2 dominates and the rocket descends.CRCR

The "correct" solution is to differentiate height to get velocity;CRCR

V(t) = d h(t)/dt = -at + 100CRCR

At the highest point the rocket changes direction so v(t)=0. Assuming that a=10 m/s^2 (gravity on Earth);CRCR

0 = 100 - at CR

t = 100 / 10 = 10 s CRCR

maximum height is; CRCR

h(t) = 100t - 0.5at^2 CR

= 1000 - 0.5 x 10 x 100 CR

= 500m CRCR

The "cheat" way to do it is to notice that the equation is specified for the time 0 < t < 20. If the rocket flight is symmetrical (true for unpowered flight in a vacuum) it will ascend for the same time that it descends and the maximum height is at the mid point t=10s. The latter is not the textbook solution but allows a quick check on your answer.

Answered by Jarvis | 5 years ago

The maximum height of the rocket will occur when it`s vertical velocity component is zero (at the apex of it`s flight path).

Find the rate of change of height h`(t) by differentiation:

h`(t) = -2t + 100 ( the formatting is not clear but I assume the first term if h(t) is -t^2 )

Find the time, t, when h`(t) = 0

-2t + 100 = 0

t = 50 seconds.

Now use the h(t) formula to find the maximum height:

h(50) = -(50)^2 + 100(50)

= 2500 m

Find the rate of change of height h`(t) by differentiation:

h`(t) = -2t + 100 ( the formatting is not clear but I assume the first term if h(t) is -t^2 )

Find the time, t, when h`(t) = 0

-2t + 100 = 0

t = 50 seconds.

Now use the h(t) formula to find the maximum height:

h(50) = -(50)^2 + 100(50)

= 2500 m

Hi, I can`t see what character you`ve put before the t2 (t-squared). Let`s call it k.

Now, h = kt^2 + 100t => dh/dt = 2kt + 100, and when the rocket reaches its maximum height, the rate of change of height with time is zero.

Therefore the relevant time t = -50/k

And now use this value of t in the original equation to work out that its height will then be

k (-50/k)^2 + 100 (-50/k) = 2500/k^2 - 5000/k

Now, h = kt^2 + 100t => dh/dt = 2kt + 100, and when the rocket reaches its maximum height, the rate of change of height with time is zero.

Therefore the relevant time t = -50/k

And now use this value of t in the original equation to work out that its height will then be

k (-50/k)^2 + 100 (-50/k) = 2500/k^2 - 5000/k

The rocket reaches its maximum height when it`s at the top of its curve, when it is momentarily still before it comes back down. when it is momentarily still, the gradient of its curve is 0, i.e. when d(h)/d(t)= 0

d(h)/d(t) = -10t + 100 = 0

10t = 100

t= 10 so it reaches its maximum height at 10 seconds after lift-off.

Put this back into the original eq to get the height:

h= -5(10)^2 + 100(10) = -500 + 1000 = 500m.

d(h)/d(t) = -10t + 100 = 0

10t = 100

t= 10 so it reaches its maximum height at 10 seconds after lift-off.

Put this back into the original eq to get the height:

h= -5(10)^2 + 100(10) = -500 + 1000 = 500m.

Answered by Madeleine | 5 years ago

you want to find the maximum height, that is when h(t) is the biggest value. the value of h increases as t increases. as t can only be a maximum of 20 seconds, to find the max height sub t=20 into the equation to get the value for h.

Q: h(t) = 5t^2 + 100t, 0 <= t <= 20,

All parts involving t are positive. The range of t is in the positive realm i.e. from 0 to 20. Therefore the rocket reaches maximum height when t is largest i.e. t=20,

Answer done in stages of working, substituting t for 20:

h(t=20) = 5(20)^2 + 100(20),

h(t=20) = 5(400) + 100(20),

h(t=20) = 2000 + 2000,

h(t=20) = 4000.

All parts involving t are positive. The range of t is in the positive realm i.e. from 0 to 20. Therefore the rocket reaches maximum height when t is largest i.e. t=20,

Answer done in stages of working, substituting t for 20:

h(t=20) = 5(20)^2 + 100(20),

h(t=20) = 5(400) + 100(20),

h(t=20) = 2000 + 2000,

h(t=20) = 4000.

This is a poorly framed question. It gives no information about what happens after 20 seconds. It is curious that the unit of distance in the question is feet: in UK Maths, for many decades, distance has been measured in metres.

The narrow answer is: within the 20-second window, the rocket reaches maximum height when t = 20 and the height at that time is 2400 feet (20^2 + 100 x 20).

However, a broader answer must consider what happens beyond the 20-second window. At t = 20, the rocket will be rising at 140 feet per second. Assuming that there is no upward acceleration after that time, and assuming deceleration due to gravity of 32 feet per second per second (and assuming that air resistance is negligible), the velocity will become zero at t = 24.375; at that time, the maximum height of the rocket`s trajectory will be reached, which is 3626.25 feet.

The narrow answer is: within the 20-second window, the rocket reaches maximum height when t = 20 and the height at that time is 2400 feet (20^2 + 100 x 20).

However, a broader answer must consider what happens beyond the 20-second window. At t = 20, the rocket will be rising at 140 feet per second. Assuming that there is no upward acceleration after that time, and assuming deceleration due to gravity of 32 feet per second per second (and assuming that air resistance is negligible), the velocity will become zero at t = 24.375; at that time, the maximum height of the rocket`s trajectory will be reached, which is 3626.25 feet.

h(t) = t2 + 100t

so dh/dt= 2t + 100

h is max when dh/dt = 0 that is, 2t+100= 0

so, t=50 but 0 ≤;; t ≤;; 20

hence t=20s when h is max

so max h = 20^2 + 100x20= 400 + 2000 = 2400ft

so dh/dt= 2t + 100

h is max when dh/dt = 0 that is, 2t+100= 0

so, t=50 but 0 ≤;; t ≤;; 20

hence t=20s when h is max

so max h = 20^2 + 100x20= 400 + 2000 = 2400ft

Cannot answer as this is what appears on my screen:

“The height h in feet of a model rocket above the ground t seconds after lift-off is given by h(t) =;;t2+ 100t, for 0 ≤;; t ≤;; 20. When does the rocket reach its maximum height above the ground? What is its maximum height?

Can you please make sure I can see the whole equation without any unreadable symbol? I suppose it is t squared, please confirm too.

Many thanks

“The height h in feet of a model rocket above the ground t seconds after lift-off is given by h(t) =;;t2+ 100t, for 0 ≤;; t ≤;; 20. When does the rocket reach its maximum height above the ground? What is its maximum height?

Can you please make sure I can see the whole equation without any unreadable symbol? I suppose it is t squared, please confirm too.

Many thanks

Summary, when finding the MAX/MIN of the dependent variable (h) in a QUADRATIC EQUATION within a RANGE on the independent variable (t), we know it occurs at either an extreme value of the independent variable (t) or at a turning point, which we find by taking the derivative and setting this equal to 0. This is possible to do without sketching a graph. However, it is a good habit to get into and will be invaluable when confronted with cubic equations.

Ignoring wind resistance and assuming the rocket returns to the ground at t=20 it will follow the path of a parabola which is symmetrical. The maximum height will therefore be when t=10. Substitute t=10 into the function to get the height. However, the function in the question cannot be correct as the number before the t^2 must be negative.

Answered by Ian | 5 years ago

This equation is not clearly written out; it looks like it is trying to be a simple differential equation, however h(t) doesn’t really make sense. Are we multiplying the height by the time? In which case the equation will need to be rearranged to get height in terms of t. Moreover is t being squared? Again the question isn’t clear. And what is the coefficient in front of t? So at present question is impossible to answer

maximum height will occur when t=20, therefore plug in 20 instead of t to achieve the max height, answer should be 2400 feet

if (h*t)=((t*2)+(100*t))

t max=20 seconds

(20*2) + (100*20)=(h*t)

((20*2)+(100*20))/20=h

h=102 feet, at either 20 seconds, if this is the point at which the rocket reaches its highest point. If however, 20 seconds is the time at which the rocket has returned back to the ground, we can presume that the rocket reaches its peak at 20/2=10 seconds.

t max=20 seconds

(20*2) + (100*20)=(h*t)

((20*2)+(100*20))/20=h

h=102 feet, at either 20 seconds, if this is the point at which the rocket reaches its highest point. If however, 20 seconds is the time at which the rocket has returned back to the ground, we can presume that the rocket reaches its peak at 20/2=10 seconds.

The maximum height is reached when the speed of the rocket is 0.

so differentiate h(t) and set this equal to 0. Then solve for t, and finally sub it back into

h(t) =t^(2)+100t for the maximum height.

I can show you the answer if you tell me what h(t) is properly.

so differentiate h(t) and set this equal to 0. Then solve for t, and finally sub it back into

h(t) =t^(2)+100t for the maximum height.

I can show you the answer if you tell me what h(t) is properly.

If you consider the graph of the function which defines the height of the rocket, it had roots at both -100 and 0. We can however eliminate any aspects of the graph which are outside the range of x is less than or equal to 0 and x is more than or equal to 20. this means that the the maximum point on the graph (of which represents the maximum height of the rocket) can be clearly seen on the graph at t = 20. the maximum height of this point is h(20) = 2400.

Considering the parabolic motion of the rocket, with the maximum height being half of the maximum time, 20 seconds, we can conclude that the rocket must reach its maximum height at 10 seconds.

Using the given function, h(10)=(10)^2+100(10), we can calculate the maximum height:

It gives us 100ft+1000ft, which equals 1,100ft.

Hence, Max height = 1,100ft at time 10 seconds.

Using the given function, h(10)=(10)^2+100(10), we can calculate the maximum height:

It gives us 100ft+1000ft, which equals 1,100ft.

Hence, Max height = 1,100ft at time 10 seconds.

Answered by Philip | 5 years ago