A projectile is projected at an angle $60$ degree with horizontal with speed $10 m / s$ . The minimum radius of curvature of the trajectory described by the projectile is (in $m$ ) :

2.55

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Solution

The correct option is A $2.55$
The radius of curvature will be minimum at the highest point of trajectory.
Horizontal velocity at that point $v_{h} = 10 cos θ = 10 \times 0.5 = 5 m / s$
Vertical velocity at that point $v_{v} = 0$
$\Rightarrow m \frac{{v_{h}}^{2}}{R_{m i n}} = m g$
$\Rightarrow R_{m i n} = \frac{{v_{h}}^{2}}{g} = \frac{(5)^{2}}{9.8} ≃ 2.55 m$

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A projectile is projected at an angle 60 degree with horizontal with speed 10m/s. The minimum radius of curvature of the trajectory described by the projectile is (in m) :

The correct option is A 2.55The radius of curvature will be minimum at the highest point of trajectory.Horizontal velocity at that point vh=10cosθ=10×0.5=5 m/sVertical velocity at that point vv=0⇒mvh2Rmin=mg⇒Rmin=vh2g=(5)29.8≃2.55m

A projectile is projected at an angle $60$ degree with horizontal with speed $10 m / s$ . The minimum radius of curvature of the trajectory described by the projectile is (in $m$ ) :