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(D) Given that the horizontal range $R$ is twice the maximum height $H$,so $R = 2H$.We know the formulas for horizontal range and maximum height are $

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

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(D) Given that the horizontal range $R$ is twice the maximum height $H$,so $R = 2H$.We know the formulas for horizontal range and maximum height are $R = \frac{u^2 \sin(2	heta)}{g}$ and $H = \frac{u^2 \sin^2 	heta}{2g}$.Substituting these into the given condition $R = 2H$:$\frac{u^2 (2 \sin 	heta \cos 	heta)}{g} = 2 \left( \frac{u^2 \sin^2 	heta}{2g} ight)$$2 \sin 	heta \cos 	heta = \sin^2 	heta$Dividing both sides by $\sin 	heta$ (assuming $\sin 	heta 
eq 0$):$2 \cos 	heta = \sin 	heta \Rightarrow 	an 	heta = 2$.From the triangle with $	an 	heta = \frac{	ext{Perpendicular}}{	ext{Base}} = \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 $R$:$R = \frac{2 u^2 \sin 	heta \cos 	heta}{g} = \frac{2 u^2}{g} \left( \frac{2}{\sqrt{5}} ight) \left( \frac{1}{\sqrt{5}} ight) = \frac{4 u^2}{5g}$.

For an object projected from the ground with speed $u$,the horizontal range is two times the maximum height attained by it. The horizontal range of the object is ..........

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$A$ projectile thrown with a speed $v$ at an angle $\theta$ has a range $R$ on the surface of the Earth. For the same $v$ and $\theta$,its range on the surface of the Moon will be:

$A$ projectile is thrown with a velocity of $50 \, m/s$ at an angle of $53^o$ with the horizontal. Determine the instants at which the projectile is at the same height.

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