Abasketball player standing up with the hoop launches the ball straight up with an initial velocity of v_o = 3.75 m/s from 2.5 m above the ground. part (a) what is the maximum height, h (in meters), above the launch point the basketball will achieve? part (b) on his first attempt, the ball doesn't make it high enough to reach the hoop. if the hoop is at 3.5 m above the ground, what is the minimum velocity (in/ms) with which be must launch his second attempt to reach the hoop?

a) The maximum height the ball will achieve above the launch point is 0.2 m.

b) The minimum velocity with which the ball must be launched is 4.43 m/s or 0.174 in/ms.

Explanation:

a)

For the height reached, we use 3rd equation of motion:

2gh = Vf² - Vo²

Here,

Vo = 3.75 m/s

Vf =  0m/s, since ball stops at the highest point

g = -9.8 m/s² (negative sign for upward motion)

h = maximum height reached by ball

therefore, eqn becomes:

2(-9.8m/s²)(h) = (0 m/s)² - (3.75 m/s²)²

h = 0.2 m

b)

To find out the initial speed to reach the hoop at height of 3.5 m, we again use 3rd eqn. of motion with h= 3.5 m - 2.5m = 1 m (taking launch point as reference), and Vo as unknown:

2(-9.8m/s²)(1 m) = (0 m/s)² - (Vo)²

(Vo)² = 19.6 m²/s²

Vo = √19.6 m²/s²

Vo = 4.43 m/s

Vo = (4.43 m/s)(1 s/1000 ms)(39.37 in/1 m)

Vo = 0.174 in/ms

a) the event horizon, singularity, and the chute located between the two.