A projectile has initially the same horizonta | vedclass

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Question

$H=\frac{u^{2} \sin ^{2} 	heta}{2 g}$

or $\quad 80=\frac{u^{2} \sin ^{2} 	heta}{2 	imes 10}$

or $\quad \mathrm{u}^{2} \sin ^{2} 	heta=1600$

$\m

Vedclass · Accepted Answer

${	an ^{ - 1}}\,\left( {\frac{4}{9}} ight)$

Vedclass · Answer

$H=\frac{u^{2} \sin ^{2} 	heta}{2 g}$

or $\quad 80=\frac{u^{2} \sin ^{2} 	heta}{2 	imes 10}$

or $\quad \mathrm{u}^{2} \sin ^{2} 	heta=1600$

$\mathbf{o r}$     $u \sin 	heta=40 \mathrm{ms}^{-1}$

Horizontal velocity $=u \cos 	heta$

$=a t$

$=3 	imes 30$

$=90 \mathrm{ms}^{-1}$

$\frac{u \sin 	heta}{u \cos 	heta}=\frac{40}{90}$

or  $	an 	heta=\frac{4}{9}$ or $	heta=	an ^{-1}\left(\frac{4}{9}ight)$

Column $I$	Column $II$
$(A)$ Average velocity between initial and final points	$(p)$ $u \sin \theta$
$(B)$ Change in velocity between initial and final points	$(q)$ $u \cos \theta$
$(C)$ Change in velocity between initial and final points	$(r)$ Zero
$(D)$ Average velocity between initial and highest points	$(s)$ None of the above

A projectile has initially the same horizontal velocity as it would acquire if it had moved from rest with uniform acceleration of $3\, ms^{-2}$ for $ 0.5\, minutes$. If the maximum height reached by it is $80\, m$, then the angle of projection is (Take $g = 10\, ms^{-2}$)