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## Homework Statement

A rocket takes off from the launch pad and moves directly upward with an acceleration of 29.4 m/s

^{2}. 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)at

^{2}

vf

^{2}= vi

^{2}+ 2ad

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 vf

^{2}= vi

^{2}+ 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)...