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Q.

A body of mass 2 kg is projected with a velocity of $40 m/s$ at an angle $60°$ with the ground. The K.E of the body at the highest point reached is

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a

Zero

b

200 J

c

300 J

d

400 J

answer is D.

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Detailed Solution

Hint: $m = 2 kg, u = {40 ms}^{- 1}, θ = 60 °$

$K.E = ?$

$K.E = \frac{1}{2} {mu}_{f}^{2}$

$K.E = \frac{1}{2} \times 2 \times u^{2} \cos^{2} 60$

$K.E = 400 J$

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