Question

a player passes a 0.600kg basketball down court for a fast break.the ball leaves the player hands with a speed of 8.30m/s and slows down to 7.20 m/s at its highest point. Ignoring air resistance,how high above the release point it is at its maximum height?

Answer 1

At the release point, the ball has kinetic energy,

$\dfrac12m{v_0}^2$

and at its maximum height it has potential and kinetic energy,

$-mgy+\dfrac12mv^2$

where $y$ is the maximum height attained above the release point.

The LoCoE tells us that we should have

$\dfrac12m{v_0}^2=-mgy+\dfrac12 mv^2$

$\implies v^2-{v_0}^2=-2gy$

$\implies\left(7.20\dfrac{\rm m}{\rm s}\right)^2-\left(8.30\dfrac{\rm m}{\rm s}\right)^2=-2\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)y$

$\implies y=0.870\,\mathrm m$

Answer 2

Answer: The maximum height of the ball is 0.87 m

Explanation:

To calculate the height of the ball, we use third equation of motion:

$v^2-u^2=2as$

where,

s = distance traveled  / height of the ball = ?

u = initial velocity of the ball = 8.30 m/s

v = final velocity of the ball = 7.20 m/s

a = acceleration due to gravity = $-9.8m/s^2$  (negative sign represents the ball is going upwards that is against gravity)

Putting values in above equation, we get:

$(7.20)^2-(8.30)^2=2\times (-9.8)\times s\\\\s=\frac{(7.20)^2-(8.30)^2}{(2\times (-9.8))}=0.87m$

Hence, the maximum height of the ball is 0.87 m