Question

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

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

net speed at highest point = ucos α

then

v = u -gt

ucos α/2 = usinα – gt

Then

$t=usin\alpha /g–ucos\alpha /2g$

$\frac{u}{2g}2sin\alpha -cos\alpha$