A particle is projected under gravity with a speed $v_{0}$ . Using $W-E$ theorem, find the speed of the particle as the function of vertical distance $y$ .

v = \sqrt{v_{0}^{2} - 2 g y}

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v = \sqrt{v_{0}^{2} - g y}

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v = \sqrt{v_{0}^{2} + 2 g y}

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v = \sqrt{v_{0}^{2} + g y}

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Solution

The correct option is A $v = \sqrt{v_{0}^{2} - 2 g y}$

Let the particle move up through a height $y$ between points 1 and 2. The work done by gravity in shifting the particle from point 1 to 2 is
$W_{gr} = - m g y$
The change in kinetic energy is
$Δ K E = \frac{1}{2} m (v^{2} - v_{0}^{2})$
Using $W-E$ theorem, $W_{n e t} = Δ K E$
i.e $\frac{m}{2} (v^{2} - v_{0}^{2}) = - m g y$
This gives $v = \sqrt{v_{0}^{2} - 2 g y}$

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