Answer
given,
speed of the basketball player, v_x = 4.75 m/s
height of the jump = ?
a) initial vertical velocity = 0 m/s
final vertical velocity = ?
height, h = 0.75 m
using equation of motion
v² = u² + 2 g h
v² = 0² + 2 x 9.8 x 0.75
v² = 14.7
v = 3.83 m/s
b) let horizontal distance= x m
maximum height at time , t s
time taken to reach maximum height
t = 0.39 s
horizontal distance
= v_x × t
= 4.75 × 0.39
= 1.85 m
horizontal distance is equal to 1.85 m