Question

A projectile is projected at an angle (a > 45) with an initial velocity u. The time t, at which its magnitude of horizontal velocity will equal of vertical velocity is

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

at time ${}^{\prime }{t}^{\prime }$

$\begin{array}{c}{V}_y=V_x=u_x=u\cos \alpha \\ \Rightarrow u\cos \alpha =u\sin \alpha -gt\\ gt=u\left(\sin \alpha -\cos \alpha \right)\\ t=\frac{u\left(\sin \alpha -\cos \alpha \right)}{g}\end{array}$

$\therefore$ Option $C$ is correct .