Physics - Kinematics Question with Solution | TestHub

PhysicsKinematics2-D & 3DMedium2 minPYQ_2021
PhysicsMediumsingle choice

The trajectory of a projectile in a vertical plane isy=αx-βx2,whereαandβare constants andx & yare respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projectionθand the maximum height attainedHare respectively given by

Options:

Answer:
D
Solution:

y=αx-βx2
comparing with trajectory equation

y=xtanθ-12gx2u2cos2θ

tanθ=αθ=tan-1α

β=12gu2cos2θ

u2=g2βcos2θ

Maximum height : H

H=u2sin2θ2 g=g2βcos2θsin2θ2 g

H=tan2θ4β=α24β

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:2-D & 3D
2mℹ️ Source: PYQ_2021

Doubts & Discussion

Loading discussions...