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,

`\frac12m{v_0}^2`

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

`-mgy+\frac12mv^2`

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

The LoCoE tells us that we should have

`\frac12m{v_0}^2=-mgy+\frac12 mv^2`

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

`\implies\left(7.20(\rm m)/(\rm s)\right)^2-\left(8.30(\rm m)/(\rm s)\right)^2=-2\left(9.80(\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

Step-by-step 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* (-9.8)* s\\\\s=((7.20)^2-(8.30)^2)/((2* (-9.8)))=0.87m`

Hence, the maximum height of the ball is 0.87 m