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

The rocket above the ground is in 44 sec.

Step-by-step 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-(1)/(2)at^2`

Put the value into the formula

`1200=92t+(1)/(2)*4t^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*10`

`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-(1)/(2)at^2`

Put the value into the formula

`0=1200+132t-(1)/(2)*9.8t^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.