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  • Initial speed of projectile fired from ground is 'u' at the maximum of motion speed of projectile is <img alt="\frac{\sqrt{3}}{5}u" src="/latex-image/?%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B5%7

Initial speed of projectile fired from ground is 'u' at the maximum of motion speed of projectile is \frac{\sqrt{3}}{5}u. find the time of the flight.

Option: 1

\mathrm{\frac{2 \sqrt{22}}{5}\left(\frac{u}{g}\right)}


Option: 2

\mathrm{\frac{2}{5}\left(\frac{u}{y}\right)}


Option: 3

\mathrm{zero}


Option: 4

None of these


Answers (1)

best_answer

At highest point vertically component of velocity zero .

\mathrm{\begin{aligned} & v \cos \theta=\frac{\sqrt{3}}{5} u \\ & \cos \theta=\frac{\sqrt{3}}{5} \\ & \sin \theta=\frac{\sqrt{22}}{5} \end{aligned}}

\mathrm{\begin{aligned} T= & \frac{2 u \sin \theta}{g}=\frac{2 \times u \times}{g} \frac{\sqrt{22}}{5} \\ & =\frac{2 \sqrt{22}}{5}\left(\frac{u}{g}\right) \end{aligned}}

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Rishabh

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