y=αx−βx2
comparing with trajectory equation
y=xtanθ−21u2cos2θgx2
tanθ=α⇒θ=tan−1α
β=21u2cos2θg
u2=2βcos2θg
Maximum height : H
H=2gu2sin2θ=2βcos2θg2gsin2θ
H=4βtan2θ=4βα2
JEE Main 2021 — Physics Mechanics
The trajectory of a projectile in a vertical plane is y=αx−βx2, where α and β are constants and x&y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection θ and the maximum height attained H are respectively given by
Held on 26 Feb 2021 · Verified 6 Jul 2026.
tan−1α,β4α2
tan−1(αβ),βα2
tan−1β,2βα2
tan−1α,4βα2
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