Question

12. A ball is thrown vertically downwards from a height
of 20 m with an initial velocity v. it collides with
the ground, loses 50 percent of its energy in
collision and rebounds to the same height. The
initial velocity v, is (Take g = 10 ms )
[Re-AIPMT-2015
(1) 10 ms 1
(2) 14 ms
(3) 20 ms1
(4) 28 ms1
elece curare a block of mass M​

Answer 1

Solution:-Let the final velocity of the ball when,it touches the ground be V1

then,

it's kinetic energy=1/2×m×(v1)^2

50 percent of kinetic energy=1/2 ×m(v1)^2×50/100=m(v1)^2/4

let the velocity after collision to the ground be u

1/2 ×m×u^2=(mv1)^2/4

u=v1/√2

when the ball rebound again,then in that case

initial velocity(u)=v1/√2

acceleration=10

final velocity(v)=0

distance travelled (s)=20m

using third law of uniformly accelerated motion

we get

0^2=u^2-2×10×20

u^2=400

u=20m/s

Now, final velocity just before before collision=u√2=20√2=20√2m/s

now,

when the body is thrown downward from velocity v

distance travelled=20

final velocity=20√2

acceleration=g

using third law we get

800=v^2+2×10×20

v=20m/s

hence it's initial velocity is 20m/s

{hope it helps you friend}