Avolleyball player bumps a ball across a net with the velocity and angle shown below. what is the maximum height of the ball?

D. 12.4 m

Explanation:

Given that,

The initial velocity of the ball, u = 18 m/s

The angle at which the ball is projected, θ = 60°

The maximum height of the ball is given by the formula

                             h = u² sin²θ/2g  m

Where,

                           g - acceleration due to gravity. (9.8 m/s)

Substituting the values in the above equation

                            h = 18² · sin²60 / 2 x 9.8

                               = 18² x 0.75 / 2 x 9.8

                               = 12.4 m

Hence, the maximum height of the ball attained, h = 12.4 m

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