Physics - Kinematics Question with Solution | TestHub

PhysicsKinematicsRelativeMedium2 minQB

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Raindrops are falling with a velocity of $102$ m/s, making an angle of $4 5^{\circ}$ with the vertical. The drops appear to be falling vertically to a man running with constant velocity. The velocity of the raindrops changes such that they now appear to be falling vertically with $3$ times the velocity they appeared earlier to the same person running with the same velocity.

After the velocity of rain drops change, the magnitude of velocity of raindrops with respect to ground is

Options:

Answer:

Solution:

$V_{R y} = 10 m / s$

$V_{m y} = 0$

Drops appear vertical to man $\Rightarrow V_{R x} = V_{m x} \Rightarrow V_{m x} = 10 m / s$

$V_{R y}^{'} = 103 m / s$

(2) $V_{R}^{'} = (V_{R x})^{2} + (V_{R y}^{'})^{2} = (103)^{2} + 1 0^{2} = 20 m / s$

(3) Final angle: $tan θ = \frac{10 m / s}{10 3 m / s} = \frac{1}{3}$ (with vertical) $\Rightarrow θ = 3 0^{\circ}$

Initial angle (with vertical) $\Rightarrow 4 5^{\circ}$

$∴$ Change $= 1 5^{\circ}$

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:Relative

⏱ 2mℹ️ Source: QB

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