Posted by Jerry on Monday, September 20, 2010 at 12:37am.
Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of 1.3 m/s parallel to the ground. Upon contact with the bat, the ball is 1 m above the ground. Payer B wishes to duplicate this bunt, in so far as he also wants to give the ball a velocity parallel to the ground and have his ball travel the same horizontal distance as player A's ball does. However, player B hits the ball when it is 1.8 m above the ground. What is the magnitude of the initial velocity that player B's ball must be given?
Does anyone know how to do this problem? I am very confused
- Physics - Anonymous, Monday, September 20, 2010 at 12:42am
well, since player b is obviously inadequate at athletics, it shows that player b is a woman, and because of this, she would not be able to hit the ball. The magnitude of the initial velocity would therefore be zero.
- Physics - rachel, Monday, September 20, 2010 at 1:04am
First you need to solve for time by using
1m=(1/2)(9.8)t^2 vertical initial velocity is 0m/s
Then you find the horizontal distance traveled by using
Then you need to find the time of player B by using
1.8m=(1/2)(9.8)(t^2) vertical initial velocity is 0
Finally to find player Bs initial horizontal velocity you use the horizontal equation
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