A ball is thrown at an angle theta and anoth | vedclass

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Question

$\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=	an ^{2} 	heta$

$\mathrm{H}_{2}-\mathrm{H}_{1}=50$

$\mathrm{H}_{2}\left(1-	an ^{2} 	hetaight)=50$

Vedclass · Accepted Answer

$15\,m,\,65\,m$

Vedclass · Answer

$\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=	an ^{2} 	heta$

$\mathrm{H}_{2}-\mathrm{H}_{1}=50$

$\mathrm{H}_{2}\left(1-	an ^{2} 	hetaight)=50$

$\Rightarrow \frac{\mathrm{u}^{2} \cos ^{2} 	heta}{2 \mathrm{g}} 	imes\left(\frac{\cos ^{2} 	heta-\sin ^{2} 	heta}{\cos ^{2} 	heta}ight)=50$

$\Rightarrow 1-2 \sin ^{2} 	heta=\frac{5}{8}$

$\Rightarrow \sin ^{2} 	heta=\frac{3}{16}$

so $\mathrm{H}_{1}=\frac{\mathrm{u}^{2} \sin ^{2} 	heta}{2 \mathrm{g}}=\frac{(40)^{2}}{2 	imes 10} 	imes \frac{3}{16}=15 \mathrm{m}$

so $\mathrm{H}_{2}=65 \mathrm{m}$

A ball is thrown at an angle $\theta $ and another ball is thrown at an angle $(90^o -\theta )$ with the horizontal from the same point with same speed $40\,ms^{-1}$. The second ball reaches $50\,m$ higher than the first ball. Find their individual heights?

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