Question

A basketball player standing under the hoop shoots the ball straight up with an initial velocity of v0Part (a) What is the maximum height, h (in meters), above the launch point the basketball will achieve?

Answer 1

Maximum height will be
`h=\sqrt{(v_0^2)/(2g)}`

Step-by-step explanation:

We have given initial velocity through which basketball is thrown
`=v_0`

Acceleration due to gravity
`=g\  m/sec^2`

At maximum height velocity will be zero

So final velocity v = 0 m /sec

According to third equation of motion
`v^2=u^2+2gh`

`0^2=v_0^2-2gh` ( Negative sign is due to upward acceleration )

`h=\sqrt{(v_0^2)/(2g)}`