Kinematics

JEE Main level, single correct, medium

257 of 308

Question 257 of 308

The trajectory of a projectile in a vertical plane is y=αx−βx2,y=\alpha x-\beta {x}^{2}, where α\alpha and β\beta 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\theta and the maximum height attained HH are respectively given by

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Answer: (D) tan⁡−1α,α24β{\tan }^{-1}\alpha ,\dfrac{{\alpha }^{2}}{4\beta }

y=αx−βx2y=\alpha x-\beta {x}^{2}
comparing with trajectory equation
y=xtan⁡θ−12gx2u2cos⁡2θy=x\tan \theta -\dfrac{1}{2}\dfrac{g{x}^{2}}{{u}^{2}{\cos }^{2}\theta }
tan⁡θ=α⇒θ=tan⁡−1α\tan \theta =\alpha \Rightarrow \theta ={\tan }^{-1}\alpha
β=12gu2cos⁡2θ\beta =\dfrac{1}{2}\dfrac{g}{{u}^{2}{\cos }^{2}\theta }
u2=g2βcos⁡2θ{u}^{2}=\dfrac{g}{2\beta {\cos }^{2}\theta }
Maximum height : HH
H=u2sin⁡2θ2g=g2βcos⁡2θsin⁡2θ2gH=\dfrac{{u}^{2}{\sin }^{2}\theta }{2g}=\dfrac{g}{2\beta {\cos }^{2}\theta }\dfrac{{\sin }^{2}\theta }{2g}
H=tan⁡2θ4β=α24βH=\dfrac{{\tan }^{2}\theta }{4\beta }=\dfrac{{\alpha }^{2}}{4\beta }

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