Phys151_PPL6-7

# Phys151_PPL6-7 - Good Serve Over the Net Over the Fence...

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Good Serve! Over the Net, Over the Fence . Continue with our discussion of two-dimensional motion as two one-dimensional motions. Q. In a volleyball game how will we know if the ball that is served can clear the net? Suppose the net is 3 m high and the server is 4 m from it. If the ball is launched from a height of 2m at an angle of 60 0 what must its speed v be for the ball to just clear the net? For that speed, where will the ball land on the opposite side? The time t to reach the net is obtained from the horizontal motion - a constant speed motion - and is 4 / v H = 4/(v cos 60 0 ) = 8/v seconds. The vertical motion is uniformly-accelerated, straight-line motion in the Earth’s field with an upward launch speed. The equation to use to reach the top of the net is y = v 0 t + 1/2 at 2 and the data includes y = -1 m, t = 8 / v s, v 0 = -v sin 60 0 and a = 10 m/s 2 . v H 3 m 4 m v v V

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Volleyball Anyone? Thus, v 0 = -0.866 v so that -1 = - 0.866 v. 8/v + 5 (8/v) 2 and therefore v 2 = 64 (5)/ (6.928 - 1). Thus, v = + 7.4 m/s. To find where the ball reaches the ground, we solve the displacement equation for the vertical motion: 2 = - 7.4 sin 60 0 (t G ) + 5 (t G ) 2 to find t G . This equation for t G is what is called a quadratic equation. There is a simple formula to extract the “roots” of a quadratic equation. Read up “Mathematical Preliminaries” and then get the correct root and substitute into the horizontal motion equation to find the answer. For fun , use the projectile demo in physicslessons.com to see where the ball lands and confirm the correctness of your numerics.
Round and Round We Go! Circular Motion Examples: merry-go-round, a spinning wheel, racing cars on a circular track. Motion repeats itself with the path or trajectory being a circle: a special two-dimensional motion. Consider uniform

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Phys151_PPL6-7 - Good Serve Over the Net Over the Fence...

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