Ball in the Air for 2.4s: Acceleration, Juggler's Speed & Time Elapsed

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The discussion focuses on the physics of a ball thrown by a juggler, which remains in the air for 2.4 seconds. The acceleration of the ball throughout its flight is consistently -9.81 m/s² while ascending and 9.81 m/s² while descending. The initial speed at which the juggler threw the ball is calculated to be 11.8 m/s, with the time taken to reach maximum height being 1.2 seconds. The maximum height achieved by the ball is 7.1 meters.

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assume that the ball was in the air for 2.4s. Answer the following questions:
a. what is the acceleration of the ball during the entire time the ball is in the air?
b. With what speed did the juggler throw the ball into the air?
c. How much time elapsed before the ball reached its maximum height?

Please..thanks
 
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a. While the ball is moving to the top (where v=0) the acceleration is -g. When it is falling it is g, where g = 9.81 metres per square second.

b. v = u + at

Total time in air = 2.4 s, hence time going up = 1.2 s. Hence, u = 0 - (-11.772) = 11.8 m/s.

c. 1.2 s

(d.) max height reached = 7.1 m.

Simply apply v^2 = u^2 + 2as. Remember to use -g for acceleration and v = 0 at max height.
 

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