Question

A 7450 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreciable air resistance. When it has reached a height of 520 m , its engines suddenly fail so that the only force acting on it is now gravity.
(a) What is the maximum height this rocket will reach above the launch pad?
(b) How much time after engine failure will elapse before the rocket comes crashing down to the launch pad?
(c) How fast will it be moving just before it crashes?

Answer 1

Answer:

a) 520m

b) 10.30 s

c) 100,95 m/s

Explanation:

a) According the given information, the rocket suddenly stops when it reach the height of 520m, because the engines fail, and then it begins the free fall.

This means the maximum height this rocket reached before falling  was 520 m.

b) As we are dealing with constant acceleration (due gravity) $g=9.8 \frac{m}{s^{2}}$ we can use the following formula:

$y=y_{o}+V_{o} t-\frac{gt^{2}}{2}$   (1)

Where:

$y_{o}=520 m$  is the initial height of the rocket (at the exact moment in which it stops due engines fail)

$y=0$  is the final height of the rocket (when it finally hits the launch pad)

$V_{o}=0$ is the initial velocity of the rocket (at the exact moment in which it stops the velocity is zero and then it begins to fall)

$g=9.8m/s^{2}$  is the acceleration due gravity

$t$ is the time it takes to the rocket to hit the launch pad

Clearing $t$:

$0=520 m+0-\frac{9.8m/s^{2} t^{2}}{2}$   (2)

$t^{2}=\frac{-520 m}{-4.9 m/s^{2}}$   (3)

$t=\sqrt{106.12 s^{2}$   (4)

$t=10.30 s$   (5)  This is the time

c) Now we need to find the final velocity $V_{f}$ for this rocket, and the following equation will be perfect to find it:

$V_{f}=V_{o}-gt$  (6)

$V_{f}=0-(9.8 m/s^{2})(10.30 s)$  (7)

$V_{f}=-100.95 m/s$  (8) This is the final velocity of the rocket. Note the negative sign indicates its direction is downwards (to the launch pad)