As horizontal velocity remains same so u cos θ = v cos α ⇒ cos θ = 2 1 ( cos α Rg ) cos α = 4 Rg cos α = 10 × 4 2 ( 2 − 1 ) × 10 × 2 1 = 2 2 2 − 1

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PhysicsKinematics2-D & 3DHard2 minQB

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An insect I crosses a cylinder of radius R with minimum jumping speed. Path of the insect is shown in figure. When insect will be just touching the cylinder at C & D its velocity will be tangent to the cylinder at that point. At point C radius of the cylinder is making an angle $α$ from the vertical. Minimum velocity of insect is u , at an angle $θ$ from the horiozntal $[$ Given : $R = \frac{2 ( 2 - 1 )}{10} m & g = 10 m / s^{2}]$

Value of $θ$ is

Options:

Answer:

Solution:

As horizontal velocity remains same so $u cos θ = v cos α$

$\Rightarrow cos θ = \frac{1}{2} (\frac{Rg}{cos α}) cos α = \frac{Rg cos α}{4} = \frac{2 ( 2 - 1 )}{10 \times 4} \times 10 \times \frac{1}{2}$

$= \frac{2 - 1}{2 2}$

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:2-D & 3D

⏱ 2mℹ️ Source: QB

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