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qz2sol_3891f10

# qz2sol_3891f10 - Δ t = 3 s Therefore the x component of...

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TA: Tomoyuki Nakayama Tuesday, September 14, 2010 PHY 2048: Physic 1, Discussion Section 3891 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 h = 0.80 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, at distance D = 40.0 m farther along the wall. a) What horizontal distance is traveled by the ball from hit to catch? The ball travels distance D in
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Unformatted text preview: Δ t = 3 s. Therefore, the x component of the initial velocity is D =v 0x Δ t ⇒ Δ t = D/ Δ t = 13.3 m/s Due to the symmetry of projectile motions, it takes 1 second the ball to reach the initial height after passing the top of the wall downward. The total flight time is t tot = 1+3+1 = 5 s The horizontal range is R = v 0x t tot = 66.5 m b) How high is the wall? By symmetry, the ball reaches the peak of the motion at time t top = t tot /2 = 2.5 s At the peak, the y component of the velocity is zero. Thus we have 0 = v 0y – gt top v 0y = gt top = 24.5 m/s The height of the wall is equal to the height of the ball at t = 1s. h = y (t = 1 s) = y + v 0y t – (1/2)gt 2 = 20.4 m...
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