Question

A particle is projected under gravity with a speed $v_0$. Using $\text{W-E}$ theorem, find the speed of the particle as the function of vertical distance $y$.

• A. $v=\sqrt{v_0^2-2gy}$
• B. $v=\sqrt{v_0^2-gy}$
• C. $v=\sqrt{v_0^2+2gy}$
• D. $v=\sqrt{v_0^2+gy}$

Answer 1

The correct option is A $v=\sqrt{v_0^2-2gy}$


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
$W_{\text{gr}}=-mgy$
The change in kinetic energy is
$\Delta KE=\frac{1}{2}m(v^2-v_0^2)$
Using $\text{W-E}$ theorem, $W_{net}=\Delta KE$
i.e $\frac{m}{2}(v^2-v_0^2)=-mgy$
This gives $v=\sqrt{v_0^2-2gy}$