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 2.70m/s2 . At 10.0s 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?

-> i keep getting 135 but that isnt the answer. heres how i did it:

displacement= (Vi)(t) + (1/2)at^2

=0+(1/2)(2.7)(10^2)

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

i got 27 for this but im not sure if its right.

Vf= 0+(10)(2.7)

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

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

e).How fast will it be moving when it does so?

Explanation / Answer

time with engines on

y = 1/2 a t^2 = 1/2 *2.7*10^2=135m
v = a t = 2.7*10 = 27 m/s

a) treat when engines stop as initial
v^2 = v0^2 + 2 a(y-y0)

0 = 27^2 + 2*-9.81*(y-135)

y=

m

b)by definition velocity at highest point = 0
c) a = g =9.81
d) downward since gravity

e) y = y0 + v0 t + 1/2 a t^2

0 = 135 +27*t+0.5*-9.81*t^2
t=8.67
but this is time after engines fail total time = 18.67 s
f)
v = v0 + at = 27-9.81*8.67=-58.0527 m/s

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