Question

A basketball player stands 15m away from the basket to make a 3-point shot. The basket is 3m (10 ft) above the ground. Assume the ball leaves the player's hands at a height of 2m, and with velocity vector Vo=(Vox, Voy). help?

a) what is the minimum initial y-direction velocity (Voy) needed for the ball to just barely reach the height of the basket?
b) in the scenario of part a), say the ball has Voy initial y-direction velocity. What should Vox be in order for the player to make the shot? (give your answer in terms of general Voy before plugging in the numbers!)
c) Would Voy from part a) be larger or smaller if the player was throwing the ball on the Moon? why?

Thank you! Much appreciated :)

Answer 1

Please see the explanation below.

Explanation:

The minimum velocity `v_oy` must be such that the distance travelled is `d=(3-2)=1m`

The acceleration due to gravity is `g=9.8ms^-2`

Apply the equation of motion

`v^2=u^2+2as`

`0=v_0y^2-2*g*d`

`0=v_0y^2-2*9.8*1`

`v_oy=sqrt(2*9.8)=4.43ms^-1`

The time taken to reach the greatest height is

`v=u+at`

`0=v_0y-g t`

`t=(v_0y)/g=4.43/9.8=0.452s`

Apply the equation

`d=v_0x*t`

As the velocity in the x-direction is constant

`v_0x=d/t=15/0.452=33.2ms^-1`

The velocity `v_0y` would be greater if the player throws the ball to the moon since the distance is greater.

`v_0y=sqrt(2gs)`