A rocket blasts off vertically from rest on the launch pad with an upward accele

Question

A rocket blasts off vertically from rest on the launch pad with an upward acceleration of 2.90 m/s2 . At 20.0s after blastoff, the engines suddenly fail, which means that the force they produce instantly stops.

(A)How high above the launch pad will the rocket eventually go?

(B)Find the rocket's velocity at its highest point.

(C)Find the magnitude of the rocket's acceleration at its highest point.

(D)Find the direction of the rocket's acceleration at its highest point.

(E)How long after it was launched will the rocket fall back to the launch pad?

(F)How fast will it be moving when it does so?

Explanation / Answer

Part A)

First, the height after the acceleration

d = vot + .5at2

d = 0 + .5(2.9)(20)2

d = 580 m

Then we need the speed at that time

vf = vo + at

vf = 0 + (2.9)(20)

vf = 58 m/s

Then we can find the distance during the slow down process

vf2 = vo2 + 2ad

0 = 582 + 2(9.8)(d)

d = 171.6 m

Total height = 171.6 + 580 = 751.6 m

Part B)

At the highest point, the velocity will be zero.

Part C)

At the highest point, only gravity is acting,so the acceleration will be 9.8 m/s2

Part D)

The direction is downward (Gravity pulls downward)

Part E)

The time during the acceleration was 20 sec

The time to get to the height...

vf = vo + at

0 = 58 + (9.8)(t)

t = 5.92 sec

The time to fall back down

d = vot + .5at2

751.6 = 0 + .5(9.8)(t2)

t = 12.4 sec

Total time = 20 + 5.92 + 12.4 = 38.3 sec

Part F)

vf = vo at

vf = 0 + (9.8)(12.4)

vf = 121.5 m/s

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