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qz1sol_3705s11 - from hit to catch The ball travels...

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TA: Tomoyuki Nakayama Monday, January 24, 2011 PHY 2048: Physic 1, Discussion Section 3705 Quiz 1 (Homework Set #2) Name: UFID: Formula sheets are not allowed. Do not store equations in your calculator. You have to solve problems on your own; memorizing final algebraic expressions from homework assignments and just plugging numbers into them will not give you full credit. Leave all your work. ________________________________________________________________________________ In the figure below right, a baseball is hit at a height h = 1.00 m and then caught at the same height. It travels alongside a wall, moving up past the top of the wall 1.20 s after it is hit and then down past the top of the wall 2.60 s later, at distance D = 40.0 m farther along the wall. a) What horizontal distance is traveled by the ball
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Unformatted text preview: from hit to catch? The ball travels distance D in Δ t = 2.60 s. Therefore, the x component of the initial velocity is D =v 0x Δ t ⇒ Δ t = D/ Δ t = 15.4 m/s Due to the symmetry of projectile motions, it takes 1.2 seconds the ball to reach the initial height after passing the top of the wall downward. Therefore, the total flight time is t tot = 1.2 + 2.6 +1.2 = 5 s The horizontal range is R = v 0x t tot = 77.0 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 = 1.2 s h = y (t = 1.2 s) = y + v 0y t – (1/2)gt 2 = 23.3 m...
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