A rocket (initially at rest) is fired vertically from the ground with a constant

Question

A rocket (initially at rest) is fired vertically from the ground with a constant upward acceleration. After 20 seconds, the engines turn off and the rocket continues upward, being slowed by gravity.  The total time between the rocket leaving the ground and reaching the top of its trajectory is 60 seconds.  (Assume air resistance can be ignored.)

a) What is the upward acceleration of the rocket?

b) What is the maximum height reached by the rocket?

CAN YOU PLEASE EXPLAIN THE ANSWER

[vshut off = 400 m/sec ,,  a = ?v/?t = 20 m/sec2 )b. total height = vave*20 + vave*40 = 12,000 m]

I have the answer, but I want you please show me how you did it

Explanation / Answer

let v is the maximum velcocity and a is the acceleration

v1 = a*t1 = 20*a

at maximum height velocity is zero

v2 = v1 - g*t2

0 = v - g*t2

0 = 20*a - 9.8*40

a = 9.8*40/20 = 19.6 m/s^2

V1 = 20*a = 392 m/s

H = 0.5*a*t1^2 + v1^2/2*g = 0.5*19.6*20^2 + 392^2/(2*9.8) = 11760 m

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