Asked by nndro
A pitching machine delivers a ball. if the ball is launched with an initial velocity of 22 m/s^2 [30d above horizontal] and the player hits it at the same height from which it was launched.
a) how long is the baseball in the air on its way to the batter?
b) how far away (range) is the pitching machine from the batter?
a) how long is the baseball in the air on its way to the batter?
b) how far away (range) is the pitching machine from the batter?
Answers
Answered by
Anonymous
u = horizontal velocity = 22 cos 30 = 19.05 m/s
d = 19.05 * T
but vertical problem same T
Vi= 22 sin 30 = 11 m/s
h= Hi + Vi T - 4.9 T^2
but h = Hi
4.9 T^2 = 11T
T = 11/4.9 = 2.24 s in air
d = 19.05 * 2.24 = 42.8 meters
d = 19.05 * T
but vertical problem same T
Vi= 22 sin 30 = 11 m/s
h= Hi + Vi T - 4.9 T^2
but h = Hi
4.9 T^2 = 11T
T = 11/4.9 = 2.24 s in air
d = 19.05 * 2.24 = 42.8 meters
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