Step-by-step explanation:
It is given that, Isaac drop ball from height of 2.0 m, and it bounces to a height of 1.5 m.
We need to find the speed before and after the ball bounce.
Let u is the initial speed of the ball when he dropped from height of 2 m. The conservation of energy holds here. So,

Let v is the final speed when it bounces to a height of 1.5 m. So,

So, the speed before and after the ball bounce is 6.26 m/s and 5.42 m/s respectively.