A boy standing in a ditch throws a baseball upward toward his father. The ball leaves his hand at ground level, with an initial speed of 16.0 m/s,at an angle of θ = 51°, from the horizontal.

Question

A boy standing in a ditch throws a baseball upward toward his father. The ball leaves his hand at ground level, with an initial speed of 16.0 m/s,at an angle of θ = 51°, from the horizontal.	
The boy's father reaches up and catches the ball over his head, at a height of 2.0 m above the ground. The father catches the ball on its way down (as shown in the Figure). Calculate how long the ball is in the air. ( g = 9.81 m/s2)

Henry · Answer

Vo = 16.0m/s[51o]
Yo = 16*sin51 = 12.43 m/s. = Vertical component of initial velocity.

Y = Yo + g*Tr = 0
Tr = -Yo/g = -12.43/-9.81 = 1.27 s. =
Rise time.

h = Yo*Tr + 0.5g*Tr^2 =
12.43*1.27 - 4.9*1.27^2 = 7.89 m. =
Max. ht.

h = ho - 0.5g*Tf^2 = 2 m
7.89 - 4.9*Tf^2 = 2
4.9Tf^2 = 7.89 - 2 = 5.89
Tf^2 = 1.20
Tf = 1.10 s. = Fall time.

Tr + Tf = 1.27 + 1.10 = 2.37 s. = Time
in air.