A ball is projected at 60^o from horizontal a | vedclass

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Question

Total momentum just before explosion

$=$ Total momentum just after explosion

$\mathrm{m} 	imes\left(200 \cos 60^{\circ}ight) \hat{\mathrm{i}}

Vedclass · Accepted Answer

$300\, m/s$ horizontal

Vedclass · Answer

Total momentum just before explosion

$=$ Total momentum just after explosion

$\mathrm{m} 	imes\left(200 \cos 60^{\circ}ight) \hat{\mathrm{i}}=\frac{\mathrm{m}}{3} 	imes 100 \hat{\mathrm{j}}+\frac{\mathrm{m}}{3} 	imes 100(-\hat{\mathrm{j}})+\frac{\mathrm{m}}{3} 	imes \overrightarrow{\mathrm{V}}$

$\mathrm{m} 	imes\left(200 \cos 60^{\circ}ight) \hat{\mathrm{i}}=\frac{\mathrm{m}}{3} 	imes \vec{\mathrm{V}}$

$\overrightarrow{\mathrm{V}}=3\left(200 \cos 60^{\circ}ight) \hat{\mathrm{i}}$

$\overrightarrow{\mathrm{V}}=300 \mathrm{m} / \mathrm{s} \hat{\mathrm{i}}$

A ball is projected at $60^o$ from horizontal at $200\, m/s$. At maximum height during its flight it explodes into $3$ equal fragments. Out of them one part travel at $100\, m/s$ vertically up while other at $100\, m/s$ vertically down, then third part will have speed just after explosion :-

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