Aball thrown straight upwards with an initial velocity of 15 m/s. calculate how long the ball will remain in the air and the maximum height the ball reaches?

The ball remain in air for 3.06 seconds.

Maximum height reached = 22.94 m

Explanation:

We have equation of motion v = u + at

At maximum height, final velocity, v =0 m/s

       Initial velocity = 15 m/s

       acceleration = -9.81 m/s²

Substituting

       0 = 15 -9.81 t

        t = 1.53 s

Time of flight = 2 x 1.53 = 3.06 s

The ball remain in air for 3.06 seconds.

We also have equation of motion v² = u² + 2as

At maximum height, final velocity, v =0 m/s

       Initial velocity = 15 m/s

       acceleration = -9.81 m/s²

Substituting

       0² = 15² - 2 x 9.81 x s

        s = 22.94 m

Maximum height reached = 22.94 m

and 1and 1

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