Question

Two particles are simultaneously projected in the same vertical plane from the same point with velocities u and v at angles $\alpha and\beta$ with horizontal. Find the time that elapses when their velocities are parallel.

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

Answer 1

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

$\overrightarrow{v_1\left(t\right)}=ucos\alpha \hat{i}+(usin\alpha -gt)\hat{j}\overrightarrow{v_2\left(t\right)}=vcos\beta \hat{i}+(vsin\beta -gt)\hat{j}when\overrightarrow{v_1\left(t\right)}\parallel \overrightarrow{v_2\left(t\right)}then\frac{ucos\alpha }{vcos\beta }=\frac{(usin\alpha -gt)}{(vsin\beta -gt)}\Rightarrow t=\frac{uvsin(\alpha -\beta )}{g(vcos\beta -ucos\alpha )}$