A particle of mass m is thrown with initial velocity v₀ at an angle $α$ with horizontal as shown in the figure. Find the work done by gravity by the time the particle climbs a point P (x,y)

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mgx

$mg \sqrt{x^{2} + y^{2}}$

$\frac{1}{2} m v_{0}^{2} - m g y$

-mgy

answer is D.

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Detailed Solution

Since the particle has moved up by y and force due to gravity is acting downward so work done by gravity will be negative. Hence the correct answer should be -mgy.

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