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The distance covered by a freely falling body in the first $3$ seconds of its motion is__

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45 m

50 m

None of the above

answer is B.

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Detailed Solution

The distance covered by a freely falling body in the first

3

second is

45 m

.
Accordsing to the second equation of motion, the distance travelled by a body in time,

(t)

having initial velocity

u

, and acceleration,

(a)

is,

S = ut + \frac{1}{2} a t^{2}

If the motion is under free fall then,
initial velocity(

u

) would be

= 0

Acceleration(

a

) will be acceleration due to gravity,

= g

S =

0 \times t + \frac{1}{2} g t^{2}

S = \frac{1}{2} g t^{2}

Given data and assumptions,
time

t = 3 \sec

g = 10 m / {s^{2}}^{}

\Rightarrow

S = \frac{1}{2} g t^{2}

\Rightarrow

S = \frac{1}{2} \times 10 \times 3^{2}

\Rightarrow

S = 45 m

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