# Q 3.10 A player throws a ball upwards with an initial speed of 29.4 ms-1.(d) To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g = 9.8 m s-2 and neglect air resistance).

Consider the motion from the instant the ball starts travelling in the upwards direction to the instant it reaches the highest point. Let the upwards direction be positive.

At the highest point, its velocity v is 0 m s-1.

Initially the velocity u is 29.4 m s-1.

Acceleration is -g = -9.8 m s-2

Using the third equation of motion we have

$\\v^{2}-u^{2}=2as \\s=\frac{v^{2}-u^{2}}{2a} \\s=\frac{0^{2}-29.4^{2}}{2\times( -9.8)} \\s=44.1\ m$

Consider the motion from the instant the ball starts travelling in the upwards direction to the instant it reaches back to the player's hand.

The displacement during this period is s = 0 m.

Initial velocity is u=29.4 m s-1

Acceleration is a=-g=-9.8 m s-2

Using the second equation of motion we have

$\\s=ut+\frac{1}{2}at^{2}\\ 0=29.t-4.9t^{2}\\ t=0\ s\\ or\\ t=6\ s$

The ball, therefore, reaches back to the player's hand in 6 seconds.

Note: The second solution of the above quadratic equation, t = 0 signifies that at the instant the ball starts travelling its displacement is 0 m.

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