A heavy stone is thrown with speed v from a c | vedclass

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(D) The speed of the stone at the moment it hits the ground can be determined using the principle of conservation of mechanical energy.According to th

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The speed does not depend on the initial direction

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(D) The speed of the stone at the moment it hits the ground can be determined using the principle of conservation of mechanical energy.According to the law of conservation of energy,the total mechanical energy at the top of the cliff must equal the total mechanical energy at the ground level.Let $m$ be the mass of the stone,$g$ be the acceleration due to gravity,$h$ be the height of the cliff,and $v$ be the initial speed.Initial energy at the top: $E_i = \frac{1}{2}mv^2 + mgh$.Final energy at the ground: $E_f = \frac{1}{2}mv_f^2$,where $v_f$ is the final speed.Since $E_i = E_f$,we have $\frac{1}{2}mv_f^2 = \frac{1}{2}mv^2 + mgh$.Simplifying this,$v_f = \sqrt{v^2 + 2gh}$.As seen from the final expression,the final speed $v_f$ depends only on the initial speed $v$,the height $h$,and the acceleration due to gravity $g$.It is independent of the angle or direction of projection. Therefore,the speed at which the stone hits the ground is the same regardless of the direction in which it is thrown.

$A$ heavy stone is thrown with speed $v$ from a cliff of height $h$. How and in which direction should it be thrown so that the stone hits the ground with maximum speed?

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