A rocket blasting off from Cape Kennedy rises above the surface of the Earth acc

Question

A rocket blasting off from Cape Kennedy rises above the surface of the Earth according to this equation:

Height (in meters) = 3 t 3

where t = the time from blastoff in seconds.

Sometime after blast off the first stage is jettisoned by explosives and falls to Earth - falling into the ocean.

If the first stage hits the ocean surface 21.22 seconds after blast off, how high was it when it was jettisoned? Because of the explosion used to separate the first stage from the rising rocket, assume that the initial velocity of the first stage was zero and that this first stage was in free fall, that is the distance it fell is given by: D = 1/2gt2.

Explanation / Answer

s = 3*t^3
v = ds/dt
a = dvdt

Let us say time taken to reach height = t
Time taken to come down , = t1

s = 1/2gt^2
3*t^3 = 1/2*9.8*t1^2
t1 = sqrt(0.612*t^3)

Total time = 21.22 s
t + sqrt(0.612*t^3) = 21.22
t = 6.93 s

High it was when it was jettisoned, S = 3*t^3
S = 3* 6.93^3 m
S = 998.4 m

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