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Unformatted text preview: P207  Fall 2008 Solutions Assignment 3 1 (a) Horizontal and vertical components of velocity are shown below. 10 20 30 40 50 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 v x [m/s] time [seconds]54321 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 v y [m/s] (b) The horizontal velocity 161km/hr = 161000/3600 = 44.7 m/s. The dis tance to the plate is 18.2m. Then the time for the ball to get half way to the plate is t half = x half v x = 18 . 3 / 2 m 44 . 7 m / s = 0.205 s (c) The horizontal velocity does not change so the time to travel the second half is the same as the time to travel the first half . (d) The total time of 0.41 s is just twice the human reaction time of . 2 s. (d) y = v y t half 1 2 gt 2 half . The initial vertical velocity v y = 0. Therefore, y = 1 2 (10 m / s 2 )(0 . 205s) 2 =0.210 m 1 (e) The change in vertical position over the second half of the horizontal distance is again. y = v y t half 1 2 gt 2 half . But this time the initial vertical velocity is not zero. v y = gt half . That is, v y is the vertical velocity gained during the first half of the trip. Then y = v y t half 1 2 gt 2 half = ( gt half ) t half 1 2 gt 2 half = 3 2 gt 2 half = 3 2 (10m / s 2 )(0 . 205m / s) 2 =0.630 m (f) The change in vertical position is greater during the second half of the trip because the initial vertical velocity for the second half of the trip is non zero . 2. (a) The vertical component of velocity at launch is v y (0) = (100m / s) sin(30) = 50m / s. The vertical velocity vs vertical position is given by v 2 y = v 2 y (0) 2 g ( y y (0)) The initial vertical position is y (0) = 0. Solving for the height y when v y = 0, we get y = v 2 y (0) 2 g = (50m / s) 2 2(10m / s 2 ) = 125 m (b) We have that the vertical component of the velocity is v 2 y = v 2 y (0) 2 gy . Therefore v y = ( v 2 y (0) 2 gy ) 1 2 We know that v y (0) = 50 m/s, and g = 10m / s 2 . The maximum height is 125 m. The horizontal component of the velocity is (100 m / s) cos(30) = 86 . 6 m / s. So fraction of maximum height y [m] v x [m/s] v...
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This note was uploaded on 10/26/2008 for the course PHYS 2207 taught by Professor Liepe, m during the Fall '07 term at Cornell University (Engineering School).
 Fall '07
 LIEPE, M
 Physics

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