A softball is hit over a third baseman's head with some speed v0 at an angle θ above the horizontal. Immediately after the ball
is hit, the third baseman turns around and begins to run at a constant velocity V=7.00m/s. He catches the ball t=2.00s later at the same height at which it left the bat. The third baseman was originally standing L=18.0m from the location at which the ball was hit. Find the angle and Vo.
1 answer:
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Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work: