A particle of mass 'm' is projected with a velocity 'u' at an angle '' with the horizontal. Work done by gravity during its descent from its highest point to, the position where its velocity vector makes an angle $\frac{α}{3}$ with the horizontal is,

see full answer

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*

An Intiative by Sri Chaitanya

$\frac{1}{2} m u^{2} \tan^{2} \frac{α}{2}$

$\frac{1}{2} m u^{2} \cos^{2} \frac{α}{2} \sin^{2} α$

$\frac{1}{2} m u^{2} \tan^{2}$ $α$

$\frac{1}{2} m u^{2} \cos^{2} α \tan^{2} \frac{α}{2}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

As horizontal Component of velocity remains constant,

work done = $Δ K E$ along vertical direction ,

$K_{y_{1}}$ = 0

$T a n \frac{α}{2} = \frac{V_{y}}{u C o s α} \Rightarrow V_{y} = u C o s α . T a n \frac{α}{2}$

$K_{y_{2}} = \frac{1}{2} m V_{y}^{2} = \frac{1}{2} m u^{2} C o s^{2} α . T a n^{2} \frac{α}{2}$

Watch 3-min video & get full concept clarity