Question

A ball is dropped from a height of 10 m . If the velocity
of the ball is reduced by 40% after striking the ground;
how much high can the ball bounce back? ​

Answer 1

Answer:1.6m

Explanation:

velocity of the ball just before striking the ground

=v²-u²=2gs

=>v²=2(9.8)(10)

=>v²=196

=>v=±14

therefore,

velocity of the ball after striking the ground

=(40% of 14)m/s

=5.6m/s

So, the ground is returning the ball at 5.6 m/s

therefore,

initial velocity(u) =5.6m/s

final velocity(v)=0(as it is going to stop after reaching a certain height)

g= -9.8m/s² (as ball is going upwards)

from the equation

v²-u²=2gs

=>0²-5.6=2(-9.8)s

=> -31.36= -19.6s

=>31.36/19.6=s

=>1.6=s

therefore, it will reach 1.6 m after striking the ground(neglecting air resistance)