1)

A ball is projected from the ground at an angle of 45° with the horizontal surface . It reaches a maximum  height of 120m and  returns to the ground. Upon hitting the ground for the first time, it loses  half of its kinetic energy . Immediately after the bounce , the velocity of the ball makes an angle of 30° with the horizontal surface . The maximum height it reaches after the bounce, in metres, is.........


A) 40

B) 25

C) 30

D) 15

Answer:

Option C

Explanation:

H=u2sin24502g=120m

   u24g=120m

  If speed  is v after the first collision, then speed should remain 12  times, as kinetic energy has reduced to half.

    v=u2

     hmax=v2sin23002g

                                       = (u2)2sin23002g

                                      = (u2/4g4)=1204=30m