# CBSE Class 11-science Answered

**A platform is moving upwards with an acceleration of 5m/s².At the moment when it's velocity is 3m/s, a ball is thrown from it with a speed of 30m/s w. r. t platform at an angle of 30 with horizontal w. r. t plat form. What is the time taken by the ball to return to the platform**

The ball is thrown with velocity u = 30 m/s at an angle 30^{o} when the platform velocity is v = 3 m/s .

Hence intial vertical component of velocity of ball is ( u sin30 + v ) = 18 m/s

Let the ball hit the platform when it reaches the height h from the position when the ball is thrown

as shown in figure.

Vertical displacement for the ball is

h = 18 t - [ (1/2) × g t^{2} ] = 18 t - 4.9 t^{2} ......................... (1)

where g = 9.8 m/s^{2} is acceleration due to gravity and t is time taken.

Platfrom is moving with acceleration a = 5 m/s^{2} . Hence vertical displacement for platform is

h = 3 t + [ (1/2) a t^{2} ] = 3t + 2.5 t^{2} ..................................(2)

Time taken t for the ball to hit the platform is calculated by equating eqn.(1) and (2)

3t + 2.5 t^{2} = 18 t - 4.9 t^{2}

From above expression , we get t ≈ 2 s