Question

A ball is dropped from a height of 10m.if the energy of the ball reduces by 40 percent after striking the ground, how high can the ball bounce back?

Answer 1

Since, height = 10 m,  Acceleration due to gravity(g) = 10 m/s^2Let the mass be mTherefore, Potential Energy= m x g x h=> Potential energy = m x 10 x 10 =100mSInce, Inital Potential energy is always equal to final kinetic energy...Therefore, Original kinetic energy= 100m  (where m is the mass)After bouncing back, Reduced Kinetic energy = (100-40)% of 100m = 60mSince, Initial kinetic energy = Final potential EnergyTherefore, 60m = m x g x h                         ( Since, mass remains constant =>60 = 10 x h                                                    and thus gets cancelled out) => h = 6 metre. Ans

Answer 2

height = 10m
gravitational acceleration = 10 m/s^2
let the mass of the object be m
gravitational potential energy = mgh = 10×10×m=100m
we know that potential energy = kinetic energy=100m
now the energy is reduced by 40 percent
the energy = (100-40) percent × 100m
=60/100×100m
= 60m
NOW, mgh = 60
=10× h = 60
= h = 60/10 (mass is constant)
= h = 6