# Gravitational Force Problem! Please help!

The gravitational force on a baseball is −Fgjˆ. A pitcher throws the baseball with velocity viˆ by uniformly accelerating it straight forward horizontally for a time interval ∆t =t–0 = t. If the ball starts from rest,
(a) through what distance does it accelerate before its release?

(b) What force does the pitcher exert on the ball?

## Answers and Replies

PhanthomJay
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Hi, Amanda. Per forum rules, you must list what you think are the relevant equations, and show an attempt at a solution, before we can provide assitance. Thanks.

Sorry!
I know that gravitational force is Fg=mg
And Vxf=Vxi + axt for constant a

PhanthomJay
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At this time, Amanda, we'd like to welcome you to Physics Forums! Ok, those 2 equations are useful. Try answering the first part of the question first,
If the ball starts from rest, through what distance does it accelerate before its release?

From your kinematic equation, you can solve for the acceleration, as long as you substitute in the correct given variables for the initial velocity of the ball starting at rest, and the final velocity of the ball as it leaves the thrower's hand. But then you'll need another one of the kimematic equations to solve for the distance. It would be a bit easier to start right off with the constant acceleration kinematic equation that relates distance with the given initial and final velocities and given time (t). But either way, what do you get for an answer for the distance the ball moves while it is still in the thrower's hands?

Ok - so if the ball starts at rest, the acceleration is zero, and the distance is zero because it is still in the pitchers hand. I think?

PhanthomJay
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Well, the ball is in the pitcher's hand from the time when it is at rest to the time the ball is released. The pitcher's hand, and thus the ball with it, moves and accelerates during that time when the ball and hand are in contact, until the ball is released with a given speed of vi. The numerical value of vi is not given, so just call it vi. Now use your equation vxf =vxi + at, where it is given that vxf = vi, vxi is 0, and t is t. Solve this equation for a. Now use another kinematic equation for distance as a function of velocities and acceleration, to solve for the distance the pitcher's arm (and the ball) moves during that acceleration period of contact.