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leslie leslie
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12 years ago
A juggler throws a ball straight up into the air. The ball remains in the air for 2.4 s before it lands back in the juggler's hand.

What is the acceleration of the ball during the entire time the ball is in the air?

With what speed did the juggler throw the ball into the air?

With what speed did the juggler how the ball reached its maximum height?

How high above the point of release did the ball rise?
How much time elapsed before the ball reached its maximum height?
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rndraonrndraon
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12 years ago
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12 years ago
1) -9.81 m/s^2 - acceleration due to the Earth's gravity.

2) v=u+at; -u=u-9.81*2.4; u=9.81*2.4/2= 11.772 m/s

3) Do not understand the question.

4) v^2=u^2+2as; 0=11.772^2-2*9.81*s; s=7.0632 m
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