Question

A projectile is fired with a velocity $u$ at angle $\theta$ with the ground surface. During the motion at any time it is making an angle $\alpha$ with the ground surface. The speed of particle at this time will be

• A. $ucos\theta sec\alpha$
• B. $ucos\theta .tan\alpha$
• C. $u^{2}co{s}^2\alpha si{n}^2\alpha$
• D. $usin\theta .sin\alpha$

Answer 1

The correct option is A $ucos\theta sec\alpha$

Speed along horizontal is $ucos\theta$ and speed along vertical is given by: $\frac{v}{ucos\theta }=tan\alpha$

Thus, velocity along y direction is $=ucos\theta tan\alpha$
Speed is $=\sqrt{{\left(ucos\theta tan\alpha \right)}^2+{\left(ucos\theta \right)}^2}=ucos\theta sec\alpha$