Question

A projectile is projected at an angle $(\alpha >{45}^{\circ })$ with an initial velocity $u.$ The time $t,$ at which its magnitude of horizontal velocity will equal the magnitude of vertical velocity is:

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

Answer 1

The correct option is A $t=\frac{u}{g}(\cos \alpha -\sin \alpha )$

$\text{Horizontal velocity in projectile}$

$\text{motion does not change because}$

$\text{there is no homizontal acceleration.}$

$V_y=u_{xy}+at$

$v_y=u\sin \alpha -gt$

$u\sin \alpha -gt=u\cos \alpha$

$u(\sin \alpha -\cos \alpha )=gt$

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

Hence, option C is correct.