A stone is thrown horizontally from the top of a 20 m high building wi

A stone is thrown horizontally from the top of a 20 m high building with a speed of 12 m/s. It hits the ground at a distance R from the building. Taking g = 10 m/s² and neglecting air resistance will give:

[amp_mcq option1=”R=12m” option2=”R=18m” option3=”R=24m” option4=”R=30m” correct=”option3″]

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UPSC NDA-1 – 2023

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The correct option is C.

For projectile motion, the horizontal and vertical components of motion are independent (neglecting air resistance). The time it takes for the stone to hit the ground is determined by the vertical distance and the acceleration due to gravity. The horizontal distance (range) is then calculated by multiplying the horizontal velocity by the time of flight.

The vertical motion is governed by $y = v_{y0}t + \frac{1}{2}gt^2$. Since the stone is thrown horizontally, $v_{y0} = 0$. The height is $y = 20$ m and $g = 10$ m/s². So, $20 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \implies 20 = 5t^2 \implies t^2 = 4 \implies t = 2$ s (time of flight). The horizontal motion is uniform with velocity $v_x = 12$ m/s. The range $R$ is given by $R = v_x \cdot t = 12 \text{ m/s} \cdot 2 \text{ s} = 24$ m.

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