A particle of mass 10kg is thrown at an angle of 37∘ with horizontal with an initial velocity of 5m/s. The kinetic energy of the projectile when it just reaches the ground will be
A
250J
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B
125J
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C
100J
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D
80J
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Solution
The correct option is B125J Using law of conservation of energy, Ki+Ui=Kf+Uf
Since, (Ui=Uf; as initial and final height of the particle is same ) ∴Ki=Kf⇒Kf=12m(u)2=12×10×52=125J