A projectile is fired with a speed u at an an | vedclass

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Question

Horizontal component of velocity $=v_{x}=u \cos 	heta$

 vertical component of velocity $v_{y}=u \sin 	heta-g t$

angle of velocity with ho

Vedclass · Accepted Answer

$u \,\,cos	heta \,\,sec\alpha$

Vedclass · Answer

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

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

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