Question

A ball is dropped from a height of 20 m. On striking
the ground, it loses 25% of its energy. To what
height will it rebound?​

Answer 1

Answer:

R.E.F image

V

2

=V

2

+2as

V=

2×10×60

=

1200

m/s

KE initial =

2

1

mv

2

=600m

KE initial =0.75 KE initial

=450m

2

1

mv

12

=450m

V

1

2

=900

V

112

=V

12

+2ah

O=900−2×10×h

h=45m

Answer 2

Answer:

15m

Explanation:

first the original total energy is mgh = m*20*10

= 200m

then the kinetic energy at the lowest point is 75% of the total energy since it loses 25% of the energy

therefore the energy at that point is 75/100 *200m

= 150m

since it rebounds, at the maximum height attained this whole energy will get converted to potential energy and hence is it equal to mgh

mgh = 1/2mv^2

mgh = 150m

m*10*h = 150m

h = 15 meters