A ball of mass 1 ;kg is thrown vertically upw | vedclass

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Question

$T _{1}=\frac{2 u _{1}}{ g }$

$T _{2}=\frac{2 u _{2} \sin 	heta}{ g }$

$T _{1}= T _{2}$

$\Rightarrow \frac{2}{ g } u _{1}=\frac{2 u _{2} \si

Vedclass · Accepted Answer

$1:1$

Vedclass · Answer

$T _{1}=\frac{2 u _{1}}{ g }$

$T _{2}=\frac{2 u _{2} \sin 	heta}{ g }$

$T _{1}= T _{2}$

$\Rightarrow \frac{2}{ g } u _{1}=\frac{2 u _{2} \sin 	heta}{ g } \Rightarrow u _{1}= u _{2} \sin 	heta$

$	ext { Height }=\frac{ u ^{2} \sin ^{2} 	heta}{2 g } \Rightarrow \frac{ H _{1}}{ H _{2}}=\frac{u_{1}^{2}}{u_{2}^{2} \sin ^{2} 	heta}$

$\Rightarrow \frac{ H _{1}}{ H _{2}}=1$

A ball of mass $1 \;kg$ is thrown vertically upwards and returns to the ground after $3\; seconds$. Another ball, thrown at $60^{\circ}$ with vertical also stays in air for the same time before it touches the ground. The ratio of the two heights are

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