Question

A rocket accelerates upward from rest at a rate of 5 m/s^2. At a height of 1000m, the engines burn out and it coasts the rest of the way with an acceleration of -9.8 m/s^2. a. What is the rocket's velocity at burnout (h=1000m) in meters/second? b. For how many seconds did the rocket engine burn? c. What is the rocket's maximum altitude?

Answer 1

Step-by-step explanation:

Given

acceleration of rocket(a)
`=5 m/s^2`

At h=1000 m rocket burn out

`v^2-u^2=2as`

`v^2-0=2* 5* 1000`

`v=√(10^4)=100 m/s`

(b) time to reach v=100 m/s

v=u+at

`100=0+5* t`

`t=(100)/(5)=20 s`

(c)Rocket maximum altitude

`v^2-u^2=2as`

here u=100 m/s

v=0

`a=9.8 m/s^2`

`s=(v^2)/(2g)`

`s=(100^2)/(2* 9.8)`

[tex]s=510.204

Therefore maximum altitude=510.204+1000=1510.204 m