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

We know that for the projectile motion with initial velocity $u$ thrown at an angle $\alpha$ with horizontal, net speed at the highest point will be only along horizontal direction and its value will be $u\cos \alpha$

Hence, half of speed at the highest point $=\frac{u\cos \alpha }{2}$.

Time taken for the vertical component of velocity to be half of the net speed at highest point can be found by the first equation of motion, $v_y=u_y+a_{y}t$, where $a_y=-g$, taking downward direction as negative, $u_y=u\sin \alpha$

$\therefore \frac{u\cos \alpha }{2}=u\sin \alpha -gt$

$t=\frac{u}{2g}(2\sin \alpha -\cos \alpha )$