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  • A projectile of mass m has Velocities  <img alt="\mathrm{3 \mathrm{~m} / \mathrm{s} \, \, and\, \, 4 \mathrm{~m} / \mathrm{s} }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%

A projectile of mass m has Velocities  \mathrm{3 \mathrm{~m} / \mathrm{s} \, \, and\, \, 4 \mathrm{~m} / \mathrm{s} }  at two points during its flight in the uniform gravitational field of the earth. If these two velocities are perpendicular to each other, then find the minimum kinetic energy of the particle during its flight?

Option: 1

\mathrm{\frac{72}{25} \mathrm{~m}}


Option: 2

\mathrm{ \frac{62}{49} \mathrm{~m}}


Option: 3

\mathrm{ \frac{72}{45} m}


Option: 4

\mathrm{ \frac{14}{15} \mathrm{~m}}


Answers (1)

best_answer


According to the given condition -

\mathrm{\begin{aligned} & 3 \cos \alpha=4 \cos (90-\alpha) \\ & 3 \cos \alpha=4 \sin \alpha \\ & \tan \alpha=\frac{3}{4} \end{aligned}}

As horizontal component of velocity at highest point is, \mathrm{ v \cos \theta=3 \cos \alpha }
l.e.

\mathrm{\begin{aligned} k_{\text {min }} & =\frac{1}{2} m(3 \cos \alpha)^2 \\ & =\frac{1}{2} m \times 3^2 \times\left(\frac{4}{5}\right)^2=\frac{72}{25} \text { Ans. } \end{aligned} }
 

Posted by

Sanket Gandhi

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