Question

John throws a baseball from the outfield from shoulder height, at an initial velocity of 29.4 m/s at an
initial angle of 30.0° with respect to the horizontal. The ball is in its trajectory for a total interval of
3.00 s before the third baseman catches it at an equal shoulder-height level. (Assume air resistance
negligible.) What is the ball's horizontal displacement?

Answer 1

Let:
u be the initial velocity,
a be its angle above the horizontal,
x be the horizontal distance,
y be the vertical distance,
t be the time,
g be theacceleration due to gravity.

Horizontally:
x = ut cos(a) ...(1)

Vertically:
y = ut sin(a) - gt^2 / 2

Putting y = 0 gives t when the ball returns to the starting height:
t(u sin(a) - gt / 2) = 0

t = 0 (initially), and on returning:
t = 2u sin(a) / g.

Substituting this value of t in (1):
x = 2u^2 sin(a) cos(a) / g
= u^2 sin(2a) / g
= 29.4^2 sin(60) / 9.81
= 76.3 m.