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(C) Let the ground be the origin $(y = 0)$ and the vertically upward direction be positive.Initial position $y_0 = 25 \; m$,initial velocity $v_0 = 20

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$5$

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(C) Let the ground be the origin $(y = 0)$ and the vertically upward direction be positive.Initial position $y_0 = 25 \; m$,initial velocity $v_0 = 20 \; m s^{-1}$,and acceleration $a = -g = -10 \; m s^{-2}$.The position of the ball at any time $t$ is given by the kinematic equation:$y = y_0 + v_0 t + \frac{1}{2} a t^2$When the ball hits the ground,$y = 0$. Substituting the values:$0 = 25 + 20t + \frac{1}{2}(-10)t^2$$0 = 25 + 20t - 5t^2$Dividing by $-5$:$t^2 - 4t - 5 = 0$Factoring the quadratic equation:$(t - 5)(t + 1) = 0$Since time cannot be negative,we take $t = 5 \; s$.

$A$ ball is thrown vertically upwards with a velocity of $20 \; m s^{-1}$ from the top of a multistorey building. The height of the point from where the ball is thrown is $25.0 \; m$ from the ground. How long will it be in $seconds$ before the ball hits the ground? Take $g = 10 \; m s^{-2}$.

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