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Physics - Kinematics Question with Solution | TestHub

PhysicsKinematicsMiscellaneousHard2 minQB

PhysicsHardsingle choice

Passage / Comprehension

Two projectiles following the same trajectory are found to be 8 m apart at the same horizontal level 2 s after the second was projected. If both projectiles were projected with the same initial horizontal velocity of 4 m/s from the same point.

The maximum height will be

Options:

Answer:

Solution:

$T_{O A} = T_{C D} = 2 s$

$T_{A B} = T_{BC} = \frac{T _{A C}}{2} = \frac{8/4}{2} = 1 s$

Time of flight $T = 2 + 1 + 1 + 2 = 6 s$

But $T = \frac{2 u _{y}}{g}$ , so $6 = \frac{2 u _{y}}{10} \Rightarrow u_{y} = 30 m / s$

Maximum height $= \frac{u _{y}^{2}}{2 g} = \frac{( 30 ) ^{2}}{2 ( 10 )} = 45 m$

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

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