Question

A basketball is bounced from a height of 1.13 m above the ground. During its descent, it converts potential energy to kinetic energy. Assuming it started its descent from rest, what is its final velocity (in m/s) upon impact with the ground?

Answer 1

try using this  (KE = 1/2*m*v²).

Answer 2

Answer:

The final velocity is 4.71 m/s

Explanation:

The potential energy corresponding to a body of mass (m) at a height (h) is given as:

$PE = Mgh-----(1)$

where g = acceleration due to gravity = 9.8 m/s2

The kinetic energy of a body of mass (m) moving at a velocity (v) is:

$KE = 1/2Mv^{2} ---------(2)$

It is given that:

the height (h) from which the basketball hits the ground = 1.13 m

Based on equation (1), the PE is:

$PE = M*9.8ms^{-2} *1.13 m = M*11.1 m^{2}s^{-2}$

During its descent, all the potential energy gets converted to kinetic energy

$PE = KE\\\\M*11.1 m^{2} s^{-2} = \frac{1}{2} *M*v^{2} \\\\v = 4.71 m/s$