Kinematics: find maximum height

  • #1

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
vf2 = vi2 + 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 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)...
 

Answers and Replies

  • #2
tnich
Homework Helper
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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.
 
  • #3
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##
 
Last edited:

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