Question

A rocket fired vertically up travels with an acceleration of 10m/s .The acceleleration ceases at the end of 60sec from the start. What is the maximum hight reached by the roket? (g=9.8m/s​

Answer 1

Answer:

The distance travelled by the rocket in 60s will be:

h1=(1/2)×10×602=18000m

h1=18km…(i)

and velocity acquired by it will be

v=10×60=600m/s...(ii)

Now, after 60 sec the rocket moves vertically up  with velocity of 600m/s and acceleration due to  gravity opposes its motion. So, it will go to a  height h2 till its velocity becomes zero such that :

0=(600)2−2gh2

or h2=18000m [as g=10m/s2]....(iii)

h2=18km

So, from Eqs. (i) and (iii) the maximum height  reached by the rocket from the ground  is:

h=h1+h2

h=18+18

h=36km

HOPE IT WILL HELP