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

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The ball's acceleration during its flight is -9.81 m/s² while ascending and 9.81 m/s² while descending. The juggler threw the ball with an initial speed of 11.8 m/s. It takes 1.2 seconds for the ball to reach its maximum height. The maximum height achieved by the ball is 7.1 meters. These calculations are based on the principles of kinematics and gravitational acceleration.
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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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