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(A) Let the initial position be $A$ and the point where the ball is caught be $B$. The highest point is $P$.Initial speed $u = 20 \,m/s$.Using the equ

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$2+\sqrt{3} \,s$

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(A) Let the initial position be $A$ and the point where the ball is caught be $B$. The highest point is $P$.Initial speed $u = 20 \,m/s$.Using the equation of motion $s = ut + \frac{1}{2}at^2$ for the displacement $s = 5 \,m$:$5 = 20t - \frac{1}{2}(10)t^2$$5 = 20t - 5t^2$Dividing by $5$:$1 = 4t - t^2$$t^2 - 4t + 1 = 0$Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$t = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} \,s$.Since the ball is caught on its way down, we take the larger time value:$t = 2 + \sqrt{3} \,s$.

$A$ ball is thrown straight upward from the ground with a speed of $20 \,m/s$. The ball is caught on its way down at a point $5 \,m$ above the ground. The time taken by the ball during the entire trip is (assume $g = 10 \,m/s^2$):

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