Answer
The speed of the ball immediately after rebounding off the wall is $~~3.7~m/s$
Work Step by Step
We can find the initial kinetic energy before striking the wall:
$K_i = \frac{1}{2}mv_i^2$
$K_i = \frac{1}{2}(0.150~kg)(5.2~m/s)^2$
$K_i = 2.028~J$
If the ball rebounds with only 50% of its initial kinetic energy, then the kinetic energy after rebounding is:
$K_f = 0.5~K_i = (0.5)(2.028~J) = 1.014~J$
We can find the speed of the ball immediately after rebounding off the wall:
$\frac{1}{2}mv_f^2 = K_f$
$v_f^2 = \frac{2K_f}{m}$
$v_f = \sqrt{\frac{2K_f}{m}}$
$v_f = \sqrt{\frac{(2)(1.014~J)}{0.150~kg}}$
$v_f = 3.7~m/s$
The speed of the ball immediately after rebounding off the wall is $~~3.7~m/s$
