A projectile is thrown with speed 40 ,ms ^{-1 | vedclass

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Question

(b)

$	an 	heta=\frac{v_y}{v_x}$

Also, $t_1+t_2=\frac{2 u \sin 	heta}{g}$

$4=\frac{2 	imes 40 	imes \sin 	heta}{10}$

$\sin 	heta=\fr

Vedclass · Accepted Answer

$	an ^{-1}\left(\frac{1}{\sqrt{3}}ight)$

Vedclass · Answer

(b)

$	an 	heta=\frac{v_y}{v_x}$

Also, $t_1+t_2=\frac{2 u \sin 	heta}{g}$

$4=\frac{2 	imes 40 	imes \sin 	heta}{10}$

$\sin 	heta=\frac{1}{2} \Rightarrow 	heta=30^{\circ}$

So, $	an 	heta=	an 30^{\circ} \Rightarrow \frac{1}{\sqrt{3}}$

$	heta=	an ^{-1}\left(\frac{1}{\sqrt{3}}ight)$

A projectile is thrown with speed $40 \,ms ^{-1}$ at angle $\theta$ from horizontal. It is found that projectile is at same height at $1 \,s$ and $3 \,s$. What is the angle of projection?

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