# ch02-p047 - 47 We neglect air resistance for the duration...

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47. We neglect air resistance for the duration of the motion (between “launching” and “landing”), so a = – g = –9.8 m/s 2 (we take downward to be the – y direction). We use the equations in Table 2-1 (with Δ y replacing Δ x ) because this is a = constant motion. (a) At the highest point the velocity of the ball vanishes. Taking y 0 = 0, we set v = 0 in vv g y 2 0 2 2 =− , and solve for the initial velocity: vg y 0 2 = . Since y = 50 m we find v 0 = 31 m/s. (b) It will be in the air from the time it leaves the ground until the time it returns to the ground ( y = 0). Applying Eq. 2-15 to the entire motion (the rise and the fall, of total time
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## This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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