Question

A projectile is fired with a speed $u$ at an angle '$\theta$' with the horizontal. Its speed when its direction of motion makes an angle '$\alpha$' with the horizontal is

• A. $ucos\theta$
• B. $ucos\theta sec\alpha$
• C. $ucos\theta sin\alpha$
• D. $ucos\theta cos\alpha$

Answer 1

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

Horizontal component of velocity $=v_x=u\cos \theta$

Vertical component of velocity $v_y=u\sin \theta -gt$

angle of velocity with horizontal $=\alpha$

$\tan a=\frac{v_y}{v_x}=\frac{\left(u\sin \theta -gt\right)}{\left(4\cos \theta \right)}$

$t=\frac{u\left(\sin \theta -\cos \theta \tan \alpha \right)}{g}$

$v_y=u\cos \theta \tan \alpha$

$v_x=u\cos \theta$

So the speed of the projectile $=\sqrt{\left({\left(v_x\right)}^2+{\left(v_y\right)}^2\right)}$

$=u\cos \theta \sec \alpha$