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(C) Given: Initial velocity $u = 29.4 \; m s^{-1}$,final velocity at maximum height $v = 0 \; m s^{-1}$,acceleration $a = -g = -9.8 \; m s^{-2}$.Using

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$44.1 \; m$ and $6 \; s$

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(C) Given: Initial velocity $u = 29.4 \; m s^{-1}$,final velocity at maximum height $v = 0 \; m s^{-1}$,acceleration $a = -g = -9.8 \; m s^{-2}$.Using the third equation of motion,$v^{2} - u^{2} = 2as$:$0^{2} - (29.4)^{2} = 2(-9.8)s$$s = \frac{-(29.4)^{2}}{-19.6} = \frac{864.36}{19.6} = 44.1 \; m$.Using the first equation of motion,$v = u + at$ to find the time of ascent $(t_{a})$:$0 = 29.4 + (-9.8)t_{a}$$t_{a} = \frac{-29.4}{-9.8} = 3 \; s$.Since the time of ascent equals the time of descent,the total time taken to return to the player's hands is $T = t_{a} + t_{d} = 3 + 3 = 6 \; s$.

$A$ player throws a ball upwards with an initial speed of $29.4 \; m s^{-1}$. To what height does the ball rise and after how long (in seconds) does the ball return to the player's hands? (Take $g = 9.8 \; m s^{-2}$ and neglect air resistance).

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