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Unformatted text preview: t p = (t 1 +t 2 )/2 = (1+4)/2 = 2.50 s At the top of a projectile motion, y component of the velocity is zero. Therefore, we get 0 = v 0ygt p v ⇒ 0y = gt p = 9.8× 2.5 = 24.5 m/s b) What horizontal distance is traveled by the ball from hit to catch? The time interval for the ball to come back to the original height from the top of the motion is the same as that for the ball to reach the top from the launching point. Thus the flight time is t = 2t p = 2 × 2.5 = 5.00 s The horizontal distance for this time interval is R = v 0x t = 20 × 5 = 100 m c) How high is the wall? The vertical displacement of the ball at t 1 = 1s is Δy(t = t 1 ) = v 0y t 1(1/2)gt 1 2 = 24.5 × 1  0.5×9.8×1 2 = 19.6 m Adding the initial height to this, we get the height of the wall. H = h + Δy(t = t 1 ) = 1.2 + 19.6 = 20.8 m...
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This note was uploaded on 01/15/2012 for the course PHY 2048 taught by Professor Field during the Fall '08 term at University of Florida.
 Fall '08
 Field
 Physics, Work

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