Question

For body thrown horizontally from the top of a tower,

Moderate

Solution

Initial vertical component of velocity is zero. Horizontal component of velocity remains unchanged.
For vertical motion ,
h = 0 x t + 1/2 gt²
⇒ t = $\sqrt{\frac{2 h}{g}}$
$\large \sqrt {\frac{{2h}}{g}}$
Range R = v x t = v $\times \sqrt{\frac{2 h}{g}}$
$\large \sqrt {\frac{{2h}}{g}}$
So option - (4) is right

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