Question

A body is projected horizontally from the top of a tower with a velocity of 10 ms^-1. If it hits the ground at an angle of 45°, then the vertical component of velocity when it hits ground (in ms^-1) is

Moderate

Solution

The horizontal component of velocity,
$v_{x} = u = 10 m s^{- 1}$
Now, $\tan θ = \tan 45 ° = \frac{v_{y}}{v_{x}} = \frac{v_{y}}{10}$
$\Rightarrow v_{y} = 10 m s^{- 1}$

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