Question

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 39.2 m/s2 . The acceleration period lasts for time 9.00 s until the fuel is exhausted. After that, the rocket is in free fall.

Find the maximum height ymax reached by the rocket. Ignore air resistance and assume a constant acceleration due to gravity equal to 9.80 m/s2 .

Answer 1

Height reached during acceleration period will be, suppose h1,

Using 2nd kinematic equation:

h1 = U*t + 0.5*a*t^2

U = initial speed of rocket = 0 m/sec

a = 39.2 m/sec^2

t = 9.00 sec

So,

h1 = 0*9 + 0.5*39.2*9.00^2

h1 = 1587.6 m

Now after this height fuel is exhausted and rocket's velocity at this point will be

V = U + a*t

V = 0 + 39.2*9.00 = 352.8 m/sec

Now rocket will move upward with gravitational acceleration reducing it's speed and at the max height speed of rocket will be zero and it will free fall downward, So height traveled during this period will be

Using 3rd kinematic equation:

V^2 = U^2 + 2*a*s

here U = initial speed of rocket at h1 = 352.8 m/sec

V = final speed of rocket at max height = 0 m/sec

a = -g = -9.81 m/sec

s = h2

So,

h2 = (0^2 - 352.8^2)/(2*(-9.81))

h2 = 6343.9 m

So max height will be

y_max = h1 + h2

y_max = 1587.6 + 6343.9

y_max = 7931.5 m = 7.93*10^3 m

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