Unformatted text preview: during the 14 s required to fall to the ground. We can use this to find the initial horizontal velocity as we did in lab. 50 m 3.57 s 14 s h h r v t r v t = = = = 16. Once again, we must remember that the vertical and horizontal motions are independent of each other. We begin by looking at the vertical motion. Because the ball loses 10 m/s in vertical speed each second, we can reason that it will take 3 s to reach the top of its path (the instantaneous speed is zero at the top). Symmetry tells us that it will also take 3 s to fall back down to the same height. Therefore, the ball is in the air for a total of 6 s. Mathematically, 2 0 30 m/s 3s = 6 s 10 m/s up v t t a ∆= = = ⇒where we chose the up direction to be positive. We now use this time to calculate the range of the ball. ( 29 ( 29 6 m s 6 s 36 m R vt = = =...
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 Fall '02
 GregoryFrancis
 Force, 6 m, 2 seconds, 10 m/s, 30 m/s, horizontal motions

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