Projection Under uniform Acceleration

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Question

A small object is projected horizontally from the top of a tower. If after 1 sec, The particle is at a height of 75 m from the ground, find its time of flight.

Moderate

Solution

After 1 sec, vertical displacement of the particle
$= \frac{1}{2} g t^{2} = \frac{1}{2} \times 10 \times {(1)}^{2} m = 5 m$
$∴ H =$ Height of tower $= (5 + 75) m = 80 m$
If T be the time of flight, then $H = \frac{1}{2} g T^{2} \Rightarrow T = \sqrt{\frac{2 \times 80}{10}} s = 4 \sec$

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