# Motion with constant acceleration

## Homework Statement

A rocket launched from the ground rises vertically with an acceleration of 30 m/s/s for 9.1 s until its motor stops. The acceleration of gravity is 9.8 m/s/s. Disregarding air resistance, what maximum height above the ground will the rocket achieve? Answer is units of km.

## Homework Equations

v = v(initial) + at
x = x(initial) + v(initial)t + (1/2)at^2

## The Attempt at a Solution

h = x - x(initial)
h = (1/2)(30)(9.1)^2
h = 1242 m
h = 1.242 km

I also tried it another way and got the same result. I don't see what could possibly be wrong.

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I actually have another problem of the same type - didn't want to create another topic.

## Homework Statement

A motorist is traveling at 12 m/s when he sees a deer in the road 43 m ahead. If the maximum negative acceleration of the vehicle is -7 m/s/s, what is the maximum reaction time (delta)t of the motorist that will allow him to avoid hitting the deer? Answer in units of s.

## Homework Equations

v = v(initial) + at
x = x(initial) + v(initial)t + (1/2)at^2

## The Attempt at a Solution

43 = 0 + 12t + (1/2)(-7)t^2
0 = -43 + 12t - (7/2)t^2
t = .457 s

the application of concepts is wrong.....

after the motor stops the rocket still has some velocity with it to carry it upward...

calculate that distance after the motor stops and then add it to get the max height..

Like guru said, you forgot its still traveling at some speed. When the rocket motor stops its not yet at its highest point. 30 m/s² for 9.1 seconds. Is an average speed of (9,1*30)/2 = 136,5 m/s.
During the rocket-powered part of the flight it travels 136,5 * 9,1 = 1242,15 m.

Now it starts decelerating. Remember, it still has alot of speed left. the speed at t=9,1 would be 9,1*30 = 273 m/s
the deceleration is 9,8 m/s². it will take about 27,857142857142857142857142857143 seconds to reach v=0. the average speed is again 136,5. 27,857142857142857142857142857143*136,5 = 3802,5 m

your now at the highest point and have traveled 3802,5 m + 1242,15 m = 5044,65 m. (5 km)

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