A person throws a ball straight up. He releases the ball at a height of 1.75 m above the ground and with a velocity of 12.0 m/s. Ignore the effects of air resistance. (a) How long until the ball reaches its highest point? (b) How high above the ground does the ball go?
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Expert's answer
2010-11-14T17:55:56-0500
When the ball reaches the highest point his velocity equals to zero: v=v0 - gt = 0 12 = 10t t = 6/5 = 1.2 s From the equation of the motion: h =h0 +v0t - gt2/2 = 1.75 + 12 * 1.2 - 10* 1.22/2 = 8.95 m.
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