p02_043

Fundamentals of Physics,Vol 1 (Chapters 1 - 20)

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43. We neglect air resistance for the duration of the motion (between “launching” and “landing”), so a = g = 9 . 8m/s 2 (we take downward to be the y direction). We use the equations in Table 2-1 (with y replacing ∆ x ) because this is a = constant motion. (a) At the highest point the velocity of the ball vanishes. Taking y 0 =0,weset v =0in v 2 = v 2 0 2 gy , and solve for the initial velocity: v 0 = 2 .S in c e y = 50 m we Fnd v 0 =31m/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 t> 0) we have y = v 0 t 1 2 gt 2 = t = 2 v 0 g which (using our result from part (a)) produces t =6 . 4 s. It is possible to obtain this without using part (a)’s result; one can Fnd the time just for the rise (from ground to highest point) from
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This document was uploaded on 04/18/2008.

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