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A boy standing at the top of a tower of $20\,m$ height drops a stone. Assuming $g = 10\,m{s^{ - 2}}$ , the velocity with which it hits the ground is:
A. $20\,m{s^{ - 1}}$
B. $40\,m{s^{ - 1}}$
C. $10\,m{s^{ - 1}}$
D. $5\,m{s^{ - 1}}$

Answer
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Hint:Initially the height of the boy from the ground is $20m$. The stone is dropped vertically downwards. We will apply the equation of motion to relate the height with velocity at which it hits the ground.

Formula used:
Newton’s equation of motion
${v^2} = {u^2} + 2gh$
Where $v = $ final velocity
$u = $ initial velocity
$h = $ height from stone is dropped

Complete step by step solution:
The sole force acting on an object or body during free fall is gravity. This external force on the object causes acceleration, which accelerates the object's velocity. Free-fall motion is therefore sometimes referred to as acceleration due to gravity in common usage. Since the stone is dropped it means the initial velocity is zero ($\therefore u = 0$).

Given in question $h = 20\,m$ height of the tower.
When there is a free fall as given in question, we can directly write the equation of motion.
${v^2} = 2gh$
\[ \Rightarrow {v^2} = 2 \times 10 \times 20\]
\[ \Rightarrow v = \sqrt {2 \times 10 \times 20} = 20\]
So, the velocity with which it hits the ground is $20\,m{s^{ - 1}}$.

Hence option A is correct.

Note: The only force driving the body's motion toward the earth is the force of attraction caused by gravity, which is the only force acting on the system because there are no other forces at play. Gravity's acceleration causes the body to move downward, hence the value of g is positive. If the situation had been exactly the opposite, with the body travelling upward, we would have assumed that g was negative.