Question

A rocket is launched to travel vertically upward with a constant velocity of $20$ m/s. After travelling for $35$ seconds, the rocket develops a snag and its fuel supply is cut off. The rocket then travels like a free body. The height achieved by it is:

Answer 1

We know that

$S=ut+\frac{1}{2}a{t}^2$

Since, it is moving at a constant velocity.

So,

$S=ut$

After $35$ seconds the distance will be

$S=20\times 35=700m$

Now.

From $700m$ above the ground now it travels at a free fall.

The initial velocity $=20m/s$

Final velocity $=0$

Acceleration due to gravity $=10$

Now,

$V^2=U^2-2gS$

$0=400-20S$

$20S=400$

$S=20m$

The total height achieved

$=700+20$

$=720m$

Time it takes to drop back to the earth is

$S=ut+0.5\times g{t}^2$

But $u=0$

So,

$720=0.5\times 10\times t^2$

$720=5t^2$

$t^2=144$

$t=12seconds$

The total time it takes is

$12+35=47seconds$

Hence, this is the answer.