If a body of mass 3 kg is dropped from the top of a tower of height 250 m, then its kinetic energy after 3 s will be

1126 J

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1048 J

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735 J

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1296.5 J

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Solution

The correct option is D $1296.5 J$
Using law of conservation of energy
Gain of kinetic energy after 3s = Decrease in potential energy
Decrease in potential = mgh ....(1)
h is the distance travelled in 3s
From second Equation of motion
$h = \frac{1}{2} g t^{2} . . . . (2)$

On substituting Equation (2) in Equation (1)

$D e c r e a s e i n p o t e n t i a l = \frac{1}{2} m g^{2} t^{2} m = 3 k g, g = 9.8 m / s^{2}, t = 3 s D e c r e a s e i n p o t e n t i a l = \frac{1}{2} \times 3 \times (9.8)^{2} \times 3^{2} = 1296.5 J Thus Kinetic energy after 3s = 1296.5$ J

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