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?

Answer 1

Answer:

A) 580m

B) 0 m/s

C) 9.8m/s^2

D) downward

E) 10.87s

F) 106.62 m/s

Explanation:

A) The distance traveled by the rocket is calculated by using the following expression:

$y=\frac{1}{2}at^2$

a: acceleration of the rocket = 2.90 m/s^2

t: time of the flight = 20.0 s

$y=\frac{1}{2}(2.90\frac{m}{s^2})(20.0s)^2=580m$

B) In the highest point the rocket has a velocity with magnitude zero v = 0m/s because there the rocket stops.

C) The engines of the rocket suddenly fails in the highest point. There, the acceleration of the rocket is due to the gravitational force, that is 9.8 m/s^2

D) The acceleration points downward

E) The time the rocket takes to return to the ground is given by:

$t=\sqrt{\frac{2y}{g}}=\sqrt{\frac{2(580m)}{9.8m/s^2}}=10.87s$

10.87 seconds

F) The velocity just before the rocket arrives to the ground is:

$v=\sqrt{2gy}=\sqrt{2(9.8m/s ^2)(580m)}=106.62\frac{m}{s}$