Question

A rocket is fired vertically upwards with initial velocity 92 m/s at the ground level. Its engines then fire and it is accelerated at 4m/s2 until it reaches an altitude of 1200 m. At that point the engines fail and the rocket goes into free-fall. Disregard air resistance.

Answer 1

Answer:

The rocket above the ground is in 44 sec.

Explanation:

Given that,

Initial velocity = 92 m/s

Acceleration = 4 m/s²

Altitude = 1200 m

Suppose, How long was the rocket above the ground?

We need to calculate the time

Using equation of motion

$s=ut-\dfrac{1}{2}at^2$

Put the value into the formula

$1200=92t+\dfrac{1}{2}\times4t^2$

$2t^2+92-1200=0$

$t=10\ sec$

We need to calculate the velocity

Using equation of motion

$v=u+at$

Put the value into the formula

$v=92+4\times10$

$v=132\ m/s$

When the rocket hits the ground,

Then, h'=0

We need to calculate the time

Using equation of motion

$h'=h+ut-\dfrac{1}{2}at^2$

Put the value into the formula

$0=1200+132t-\dfrac{1}{2}\times9.8t^2$

$4.9t^2-132t-1200=0$

$t=34\ sec$

When the rocket is in the air it is the sum of the time when it reaches 1000 m and the time when it hits the ground

So, the total time will be

$t'=34+10$

$t'=44\ sec$

Hence, The rocket above the ground is in 44 sec.