Answer:
We need to find height of the ball above hand = ΔX = ?
Initial velocity Vi is given = 15 m/s in the upward direction
Gravitational acceleration = a = g = 9.81 m/s² downward direction
Time to reach to the height ΔX = T = 2.5 s
Vf is not given
Second equation of motion

ΔX = 6.84 m
Step-by-step explanation:
Given data:
Vi = 15 m/s
T = 2.5 s
ΔX = ?
Using second equation of motion
ΔX = ViT - 0.5gT²
ΔX = (15)(2.5) - (0.5)(9.81)(2.5)²
ΔX = 37.5 - 30.66
ΔX = 6.84 m