A shell is fired vertically upwards with a velocity $v_{1}$ from the deck of a ship travelling at a speed of $v_{2}$ . A person on the shore observes the motion of the shell as parabola. Its horizontal range is given by:

\frac{2 v_{1}^{2} v_{2}}{g}

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\frac{2 v_{1} v_{2}^{2}}{g}

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\frac{2 v_{1} v_{2}}{g}

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\frac{2 v_{1}^{2} v_{2}^{2}}{g}

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Solution

The correct option is C $\frac{2 v_{1} v_{2}}{g}$

Time taken by shell to return to ground $= t = 2 * (\frac{v_{1} - 0}{g}) = \frac{{2 v}_{1}}{g}$

Range will be equal to the distance traveled in same time $= v_{2} t = \frac{2 v_{1} v_{2}}{g}$

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