A pitched ball is hit by a batter at a 50° angle and just clears the outfield fence, 97 m away. Assume that the fence is at the same height as the pitch and find the velocity of the ball when it left the bat. Neglect air resistance.

Question

whats the answer
m/s at 50 degrees

Answer 1

t is time to top of parabola
u is constant horizontal speed
v is variable vertical speed
S is speed leaving the bat
then
u = S cos 50
top of arc is 97/2 from bat
so
97/2 = u t
and t = 97/(2 S cos 50)
vi = initial v = S sin 50
v = vi - g t = S sin 50 - 9.8 t
at top, 97/2 meters away, v = 0
0 = S sin 50 - 9.8 (97/(2 S cos 50))
solve for S