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Question

(d)

$R=24$ also,$\frac{H}{R}=\frac{1}{4} 	an 	heta$

$\frac{H}{R}=\frac{1}{2} \Rightarrow \frac{1}{2}=\frac{1}{4} 	an 	heta$

$	an 	heta=

Vedclass · Accepted Answer

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

Vedclass · Answer

(d)

$R=24$ also,$\frac{H}{R}=\frac{1}{4} 	an 	heta$

$\frac{H}{R}=\frac{1}{2} \Rightarrow \frac{1}{2}=\frac{1}{4} 	an 	heta$

$	an 	heta=2=\frac{P}{B}$

$R=\frac{2 u^2 \sin 	heta \cos 	heta}{g}$

$R=\frac{2 u^2}{g} \cdot \frac{2}{\sqrt{5}} 	imes \frac{1}{\sqrt{5}}$

$R=\frac{4 u^2}{5 g}$

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

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