Physics - Kinematics Question with Solution | TestHub

PhysicsKinematicsRelativeMedium2 minQB

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Passage / Comprehension

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.

The magnitude of velocity of man 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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