Asked by Gogol
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.
whats the answer
m/s at 50 degrees
whats the answer
m/s at 50 degrees
Answers
Answered by
Damon
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
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
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