A particle is projected with a velocity v suc | vedclass

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Question

(A) Let the velocity of projection be $v$ and the angle of projection be $	heta$.Given that the range $R$ is twice the maximum height $H$,i.e.,$R = 2

Vedclass · Accepted Answer

$\frac{4 v^2}{5 g}$

Vedclass · Answer

(A) Let the velocity of projection be $v$ and the angle of projection be $	heta$.Given that the range $R$ is twice the maximum height $H$,i.e.,$R = 2H$.We know the formulas for range and maximum height are:$R = \frac{v^2 \sin 2	heta}{g} = \frac{2v^2 \sin 	heta \cos 	heta}{g}$$H = \frac{v^2 \sin^2 	heta}{2g}$Substituting these into the given condition $R = 2H$:$\frac{2v^2 \sin 	heta \cos 	heta}{g} = 2 \left( \frac{v^2 \sin^2 	heta}{2g} ight)$$\frac{2v^2 \sin 	heta \cos 	heta}{g} = \frac{v^2 \sin^2 	heta}{g}$$2 \cos 	heta = \sin 	heta \Rightarrow 	an 	heta = 2$.From the triangle with $	an 	heta = \frac{	ext{opposite}}{	ext{adjacent}} = \frac{2}{1}$,the hypotenuse is $\sqrt{2^2 + 1^2} = \sqrt{5}$.Thus,$\sin 	heta = \frac{2}{\sqrt{5}}$ and $\cos 	heta = \frac{1}{\sqrt{5}}$.Now,calculate the range $R$:$R = \frac{2v^2 \sin 	heta \cos 	heta}{g} = \frac{2v^2}{g} \left( \frac{2}{\sqrt{5}} ight) \left( \frac{1}{\sqrt{5}} ight) = \frac{4v^2}{5g}$.

Column $I$	Column $II$
$(A)$ Relative acceleration between two	$(p)$ $0$ $SI$ unit
$(B)$ Relative velocity between two	$(q)$ $5$ $SI$ unit
$(C)$ Horizontal distance between two	$(r)$ $10$ $SI$ unit
$(D)$ Vertical distance between two	$(s)$ $20$ $SI$ unit

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 highest points	$(r)$ Zero
$(D)$ Average velocity between initial and highest points	$(s)$ None of the above

$A$ particle is projected with a velocity $v$ such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is ($g=$ acceleration due to gravity).

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