Question

A particle is projected with a velocity $u$, at an angle $\alpha$ with the horizontal. Time at which its vertical component of velocity becomes half of its net speed at the highest point will be

• A. $\frac{u}{2g}$
• B. $\frac{u}{2g}(\sin \alpha -\cos \alpha )$
• C. $\frac{u}{2g}(2\cos \alpha -\sin \alpha )$
• D. $\frac{u}{2g}(2\sin \alpha -\cos \alpha )$

Answer 1

The correct option is D $\frac{u}{2g}(2\sin \alpha -\cos \alpha )$

Half of speed at highest point $=\frac{u\cos \alpha }{2}\therefore \frac{u\cos \alpha }{2}=u\sin \alpha -gt$
$t=\frac{u}{2g}(2\sin \alpha -\cos \alpha )$
$\therefore$ (d)