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(D) The maximum horizontal range $R_{\max}$ for a projectile is given by $R_{\max} = \frac{u^2}{g}$.Given $R_{\max} = 80 \,m$, we have $80 = \frac{u^2

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$40$

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(D) The maximum horizontal range $R_{\max}$ for a projectile is given by $R_{\max} = \frac{u^2}{g}$.Given $R_{\max} = 80 \,m$, we have $80 = \frac{u^2}{g}$, which implies $u^2 = 80g$.When the ball is thrown vertically upward with the same velocity $u$, the maximum height $H$ reached is given by the kinematic equation $v^2 = u^2 - 2gH$.At maximum height, the final velocity $v = 0$.Substituting the values: $0 = u^2 - 2gH$.$2gH = u^2$.Since $u^2 = 80g$, we get $2gH = 80g$.$H = \frac{80g}{2g} = 40 \,m$.

$A$ player can throw a ball to a maximum horizontal distance of $80 \,m$. If he throws the ball vertically with the same velocity, then the maximum height reached by the ball is (in $\,m$)

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