recall that the range of such a projectile is
R = v^2/g sin2θ
so plug in your numbers and solve for v
A pitched ball is hit by a batter at a 40° angle and just clears the outfield fence, 103 m away. If the fence is at the same height as the pitch, find the velocity of the ball when it left the bat. Ignore air resistance.
2 answers
Initial speed up = Vi = S sin 40 = 0.643 S
initial (and final) horizontal speed = u = S cos 40 = 0.766 S
Horizontal problem:
103 = 0.766 S t where t is time in air
Vertical problem:
h = 0 = Vi t - (1/2) (9.81) t^2
0 =0 0.643 S t - 4.9 t^2 = t ( .643 S - 4.9t )
well we all know the height is 0 when t is 0 so we need
.643 S = 4.9 t
but we know 103 = 0.766 S t from the horizontal problem so
.643 S = 4.9 [ 103 / ( 0.766S) ]
0.493 S^2 = 505
S = 32 meters / second at 40 deg above horizontal
initial (and final) horizontal speed = u = S cos 40 = 0.766 S
Horizontal problem:
103 = 0.766 S t where t is time in air
Vertical problem:
h = 0 = Vi t - (1/2) (9.81) t^2
0 =0 0.643 S t - 4.9 t^2 = t ( .643 S - 4.9t )
well we all know the height is 0 when t is 0 so we need
.643 S = 4.9 t
but we know 103 = 0.766 S t from the horizontal problem so
.643 S = 4.9 [ 103 / ( 0.766S) ]
0.493 S^2 = 505
S = 32 meters / second at 40 deg above horizontal