If a person can throw a stone to a maximum he | vedclass

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(C) When a stone is thrown vertically upward with initial velocity $u$, the maximum height reached is given by the formula:$h = \frac{u^2}{2g}$From th

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$2h$

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(C) When a stone is thrown vertically upward with initial velocity $u$, the maximum height reached is given by the formula:$h = \frac{u^2}{2g}$From this, we can express the square of the initial velocity as:$u^2 = 2gh$When the same stone is thrown horizontally (as a projectile) to achieve the maximum horizontal range $R_{\max}$, the angle of projection should be $	heta = 45^{\circ}$. The formula for maximum horizontal range is:$R_{\max} = \frac{u^2}{g}$Substituting the value of $u^2$ from the first equation into the second equation:$R_{\max} = \frac{2gh}{g}$$R_{\max} = 2h$Thus, the maximum horizontal distance the person can throw the stone is $2h$.

If a person can throw a stone to a maximum height of $h$ meters vertically, then the maximum distance through which it can be thrown horizontally by the same person is

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