Question

A player passes a 0.600-kg basketball downcourt for a fast break. The ball leaves the player’s hands with a speed of 8.30 m/s and slows down to 7.10 m/s at its highest point.

a. Ignoring air resistance, how high above the release point is the ball when it is at its maximum height?
b. How would doubling the ball’s mass affect the result in part a.? Explain.

Answer 1

Step-by-step explanation:

Decrease in kinetic energy of the ball

= 1/2 m ( v² - u² )

= 1/2 x .6 ( 8.3² - 7.1² )

= 5.544 J

This energy will be converted into potential energy

mgH = 5.544 ( H is maximum height attained )

H = 5.544 /( .6 x 9.8 )

= .942 m

= 94.2 cm

b ) In a ) we used the relation

mgH = 1/2 m ( v² - u² )

H = ( 1/2g ) x ( v² - u² )

Here we see H or height attained does not depend upon mass as m cancels out on both sides.