STD 11 - 3.2. motion in plane — NEET STD 11 Science — Question

 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 $=\sqrt{\left(\left(v_{x}\right)^{2}+\left(v_{y}\right)^{2}\right)}$

 $=u \cos \theta \sec \alpha$