qz2sol_3705sp10

# qz2sol_3705sp10 - t p = (t 1 +t 2 )/2 = (1+4)/2 = 2.50 s At...

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TA: Tomoyuki Nakayama Monday, January 25 th , 2010 PHY 2048: Physic 1, Discussion Section 3705 Quiz 2 (Homework Set #3) Name: UFID: Formula sheets are not allowed. Calculators are allowed. Do not store equations in your calculator. You need to show all of your work for full credit. ________________________________________________________________________________ In the figure below right, a baseball is hit at a height of h = 1.20 m and then caught at the same height. It travels alongside a wall, moving up past the top of the wall 1.00 s after it is hit and then down past the top of the wall 3.00 s later (4.00 s after the ball is hit) at distance D = 60.0 m father along the wall. a) What are the horizontal and vertical components of the ball just after being hit? The ball travels 60 m horizontally in 3 s. We have D = v 0x Δt v 0x = D/Δt = 20.0 m/s The ball reaches its peak of the motion at
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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 0y-gt 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.

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