Answer:
The speed with which the baseball leaves the hand = 20.58 m/s
Step-by-step explanation:
The time take to reach highest height during a projectile's flight is given by
t = (u sin θ)/g
u = initial velocity of the baseball = ?
θ = angle of throw above the horizontal
g = acceleration due to gravity = 9.8 m/s²
1.05 = (u sin 30)/9.8
u = (1.05 × 9.8)/0.5
u = 20.58 m/s