ch04-p042 - 42. In this projectile motion problem, we have...

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(b) Since 22 0 (19 m/s) 31 m/s y v += (the first point on the graph), we find 0 24.5 m/s. y v = Thus, with t = 2.5 s, we can use 2 1 max 0 0 2 y yy v t g t −= − or vv g y y y y 2 0 2 0 02 == − max , bg or () 1 max 0 0 2 y y y yv v t −= + to solve. Here we will use the latter: max 0 0 max 11 ( ) (0 24.5m/s)(2.5 s) 31 m v v t y + ¡ =+ = where we have taken y 0 = 0 as the ground level. 42. In this projectile motion problem, we have v 0 = v x = constant, and what is plotted is
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