A ball is projected from top of a tower with a velocity of 5 m/s at an angle of 53° to horizontal. Its speed when it is at a height of 0.45 m from the point of projection is
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A ball is projected from top of a tower with a velocity of 5 m/s at an angle of 53° to horizontal. Its speed when it is at a height of 0.45 m from the point of projection is

G1 Physics · Accepted Answer

Mechanical energy gives v² = u² - 2gΔy. Here u = 5 m/s and the particle is 0.45 m above the launch point, so v² = 25 - 2(10)(0.45) = 16. Hence v = 4 m/s, option C.