ch11-p011

# ch11-p011 - 11. To find where the ball lands, we need to...

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position for this part of the problem) and take + x rightward and + y downward. Then (since the initial velocity is purely horizontal) the projectile motion equations become xv t y g t == and 1 2 2 . Solving for x at the time when y = h , the second equation gives th g = 2. Then, substituting this into the first equation, we find () 2 2 2.0 m 2 7.48 m/s 4.8 m. 9.8 m/s h g = 11. To find where the ball lands, we need to know its speed as it leaves the track (using conservation of energy). Its initial kinetic energy is K i = 0 and its initial potential energy is U i = M gH . Its final kinetic energy (as it leaves the track) is KM v I f =+ 1 2 2 1 2 2 ω (Eq. 11-5) and its final potential energy is M gh . Here we use v to denote the speed of its
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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