A projectile is thrown into a space so as to have maximum horizontal range $R$ . Taking the point of projection as origin, find out the coordinates of the point where the speed of the particle is minimum.

(R / 2, R / 4) .

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(R / 3, R / 4) .

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(R / 2, R / 2) .

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(R / 3, R / 2) .

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Solution

The correct option is A $(R / 2, R / 4) .$
$R = \frac{u^{2} sin 2 θ}{g} = \frac{u^{2}}{g}$
Speed of particle is minimum at highest point
$H = \frac{u^{2} {sin}^{2} θ}{g} = \frac{u^{2}}{4 g} = R / 4$
At $R / 2$ in horizontal direction attain highest point.
$(R / 2), (R / 4)$

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A projectile is thrown into a space so as to have maximum horizontal range R. Taking the point of projection as origin, find out the coordinates of the point where the speed of the particle is minimum.

The correct option is A (R/2,R/4).R=u2sin2θg=u2gSpeed of particle is minimum at highest pointH=u2sin2θg=u24g=R/4At R/2 in horizontal direction attain highest point.(R/2),(R/4)

A projectile is thrown into a space so as to have maximum horizontal range $R$ . Taking the point of projection as origin, find out the coordinates of the point where the speed of the particle is minimum.

The correct option is A $(R / 2, R / 4) .$
$R = \frac{u^{2} sin 2 θ}{g} = \frac{u^{2}}{g}$
Speed of particle is minimum at highest point
$H = \frac{u^{2} {sin}^{2} θ}{g} = \frac{u^{2}}{4 g} = R / 4$
At $R / 2$ in horizontal direction attain highest point.
$(R / 2), (R / 4)$