A ball is projected with a velocity 5 text{ m | vedclass

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(B) Given, initial velocity $u = 5 	ext{ m/s}$.Horizontal range $R$ is twice the maximum height $H$, i.e., $R = 2H$.We know that $R = \frac{u^2 \sin

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given, initial velocity $u = 5 	ext{ m/s}$.Horizontal range $R$ is twice the maximum height $H$, i.e., $R = 2H$.We know that $R = \frac{u^2 \sin 2	heta}{g}$ and $H = \frac{u^2 \sin^2 	heta}{2g}$.Substituting these into the given condition:$\frac{u^2 \sin 2	heta}{g} = 2 \left( \frac{u^2 \sin^2 	heta}{2g} ight)$$\frac{u^2 (2 \sin 	heta \cos 	heta)}{g} = \frac{u^2 \sin^2 	heta}{g}$$2 \sin 	heta \cos 	heta = \sin^2 	heta$$2 \cos 	heta = \sin 	heta \Rightarrow 	an 	heta = 2$.For $	an 	heta = 2$, we have a right-angled triangle with perpendicular $= 2$ and base $= 1$. The hypotenuse is $\sqrt{2^2 + 1^2} = \sqrt{5}$.Thus, $\sin 	heta = \frac{2}{\sqrt{5}}$ and $\cos 	heta = \frac{1}{\sqrt{5}}$.Now, calculate the range $R$:$R = \frac{u^2 (2 \sin 	heta \cos 	heta)}{g} = \frac{5^2 	imes 2 	imes (\frac{2}{\sqrt{5}}) 	imes (\frac{1}{\sqrt{5}})}{10}$$R = \frac{25 	imes 2 	imes 2}{10 	imes 5} = \frac{100}{50} = 2 	ext{ m}$.Therefore, the correct option is $B$.

$A$ ball is projected with a velocity $5 \text{ m/s}$, such that its horizontal range is twice the greatest height attained. The value of the range is (in $\text{ m}$)

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