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?
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)...