ps2b4 - Ph 111.01 Fall, 2008 Solutions, Problem set 2b (4)...

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Ph 111.01 Fall, 2008 Solutions, Problem set 2b (4) Problem 1 . A batter pops a ball straight up. (a) If the ball returns to the height from which it was hit [4.3] s later, what was its initial speed? We can find the time from 2 1 00 2 x xv t a t =+ + Here the initial position x 0 and the final position x are both equal to zero, and we know a = -9.8 m/s 2 and t = 4.3 s. So, () 2 1 0 2 2 1 0 2 2 11 0 22 9.8 m/s 4.3 s 21.1 m/s vt g t g t vg t =+ − = == = (b) Does the ball reach its maximum height [2.15] s after it was hit, more than [2.15] s after it was hit, or less than [2.15] s after it was hit? Explain. 2.15 s is half of the total time. Coming down is just the reverse of going up, so the ball reaches its maximum height just 2.15 s after being hit. (c) Find the maximum height of the ball, as measured from the point where it was hit. We can use the equation we started with above, and put in t = 2.15 s, the time when the ball is at its maximum height. ( ) 2 1 0 2 2 2 1 2 0 21.1 m/s 2.15 s 9.8 m/s 2.15 s 22.7 m hv t a t =+ + =− = Problem 2. An expectant father paces back and forth, producing the position-versus-time graph shown in Figure 2-29. (The horizontal axis is marked in increments of
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This note was uploaded on 05/13/2010 for the course PHYSICS 111 taught by Professor Staff during the Fall '08 term at S.F. State.

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ps2b4 - Ph 111.01 Fall, 2008 Solutions, Problem set 2b (4)...

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