# Kinematics: find maximum height

## Homework Statement

A rocket takes off from the launch pad and moves directly upward with an acceleration of 29.4 m/s2. It runs out of fuel after 4s and continues to coast upward, reaching a maximum height before falling back down to Earth.

a) Find the rocket's maximum height.
b) What is the velocity the instant before the rocket crashes back to the ground?

## Homework Equations

d = vi*t + (1/2)at2
vf = vi+at
a = v/t

## The Attempt at a Solution

Maximum height is when the final velocity is zero, so:
vf = 0 m/s, vi = 0m/s, a = 29.4 m/s^2

Using vf2 = vi2 + 2ad
0 = 0 + 2*29.4d
0 = 58.8d
d = 0

a) I tried a=v/t and get a time of zero when rearranging it as well. This doesn't make any sense, how is the time zero and distance 0 at maximum height? I know this is likely meaning d and t = 0 at the beginning of the flight, but how can I find the max height?

b) This sounds like instantaneous velocity or something, but I searched up the formula and it requires calculus, and our class isn't supposed to have calc in it. So I assume I have to use a kinematic equation, but I'm not sure which (either vf2 = vi2+2ad or vf=vi+at)...

tnich
Homework Helper
First, draw a picture of the rocket's trajectory. Label all of the important points with height, acceleration, velocity and time. So for example, at time ##t=0##, ##h=0##, ##a=29.4##, and ##v = 0##. How about at time ##t=4##. (Hint: ##h = h_4##, . . .). You also need to do this when the height is at a maximum, and when the rocket hits the ground. If you don't know the values of ##t##, ##h##, ##a##, and ##v##, just put labels on them (##h=h_{max}## at the maximum height, for example). Then write your equations.

After you have calculated the velocity at t=4s, I think is better calculate the height using energy with the Newton's formula: ##U=-G\frac{M·m}{R}## , (##m## cancels in the calculation), because we don't know if with these velocity g is a constant during all the trajectory.
Also for the go back.
Edit.
I have calculated the initial velocity afther the acceleration and perhaps you can use the simplify formula for potential energy ##U=m·g·h##

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