Projection Under uniform Acceleration

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Question

A projectile is projected at an angle of $60^{0}$ with the horizontal with speed of 10 m/s. The minimum radius of curvature of the trajectory described by the projectile is

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Solution

The radius of curvature is minimum at the highest point where the centripetal force = mg And speed V = $u \cos α$
Now $\frac{{mV}^{2}}{ρ} = mg$
$\Rightarrow \frac{{(ucosα)}^{2}}{ρ} = g \Rightarrow ρ = u^{2} \cos^{2} α / g \Rightarrow ρ_{\min} = \frac{{(ucosα)}^{2}}{g} = \frac{{(10 \cos 60^{0})}^{2}}{9 .8} m = 2 .55 m .$

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Projection Under uniform Acceleration

Remember concepts with our Masterclasses.

A projectile is projected at an angle of 600 with the horizontal with speed of 10 m/s. The minimum radius of curvature of the trajectory described by the projectile is

The radius of curvature is minimum at the highest point where the centripetal force = mg And speed V = u cos αNow mV2ρ=mg⇒ucosα2ρ=g ⇒ρ=u2cos2α/g ⇒ ρmin = ucosα2g = 10cos60029.8 m=2.55 m.

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A projectile is projected at an angle of $60^{0}$ with the horizontal with speed of 10 m/s. The minimum radius of curvature of the trajectory described by the projectile is