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The gravitational force exerted on a baseball is 2.31 N down. A pitcher throws the ball horizontally with velo?
The gravitational force exerted on a baseball is 2.31 N down. A pitcher throws the ball horizontally with velocity 14.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 152 ms. The ball starts from rest.
(a) Through what distance does it move before its release?
(m)
(b) What are the magnitude and direction of the force the pitcher exerts on the ball?
magnitude (N)
direction (° above the horizontal)
1 Answer
- Anonymous9 years agoFavorite Answer
OK, for (a) first, find the acceleration-- a=change in v/change in time.
18m/.152s=118.42m/s.
Now, plug this into the equation x=Xo+VoyT+.5(a)(t)^2. You are trying to find the total displacement during the throw. Since Xo and Vo are both zero, you get .5(118.42)(.152)^2=1.36m!
b) the last part is simple, remember F=MA. So find the mass (2.31N/9.8m/s^2) which is .235 kg. Now multiply this by the acceleration of 118.42 m/s^2. The answer is 27.9 N.
However, I don't know how to find the acceleration...