A rocket, initially at rest on the ground, accelerates straight upward from rest

Question

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant net acceleration a, until time t1, when the fuel is exhausted. Find the maximum height H that the rocket reaches (neglecting air resistance). Express the maximum height in terms of a, t1 , and/or g. Note that in this problem, g is a positive number equal to the magnitude of the acceleration due to gravity. The correct answer involves the variable g , which was not part of your answer. If the rocket's net acceleration is a = 3g for t1 = 5.00 s, what is the maximum height the rocket will reach? Express your answer numerically in meters, using g = 9.81 m/s2.

Explanation / Answer

d=do+vot+1/2at^2
if you take the point where it starts at the ground and then ends at the ground, you get
0=vot+1/2at^2
so
vo=1/2gt
since a=-9.8 it becomes g (which is 9.8) when you move it to the other side

knowing that, now use that equation for the rocket starts at the ground and the final is when it is at maximum height

d=vo(t/2)+1/2a(t/2)^2
substitute the intial velocity found earlier and you get
d=1/2gt (t/2) + 1/2a(t/2)^2
simplify and get
d=1/4gt^2+(1/16)at^2
d=1/4gt^2(1-1/4)
d=3/16gt^2

Hmax =(1/2)*(3*9.81)(5^2) +(3*9.81*5)^2/(2*9.81)

Hmax =1471.5 m

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