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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
$u\,\, sec\theta \,\,cos\alpha$
$u\,\, sec\theta\,\, sin\alpha$
$u \,\,cos\theta \,\,sec\alpha$
$u\,\, sin \theta \,\,sec\alpha$
Solution
Horizontal component of velocity $=v_{x}=u \cos \theta$
vertical component of velocity $v_{y}=u \sin \theta-g t$
angle of velocity with horizontal $=\alpha$
$\tan a=\frac{v_{y}}{v_{x}}=\frac{(u \sin \theta-g t)}{(4 \cos \theta)}$
$t=\frac{u(\sin \theta-\cos \theta \tan \alpha)}{g}$
$v_{y}=u \cos \theta \tan \alpha$
$v_{x}=u \cos \theta$
So the speed of the projectile $=\sqrt{\left(\left(v_{x}\right)^{2}+\left(v_{y}\right)^{2}\right)}$
$=u \cos \theta \sec \alpha$
