JEE PHYSICS MCQ: MOTION IN PLANE

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Application +4 | -1

The trajectory of a projectile in a vertical plane is

y = \alpha x - \beta {x^2}

, where

\alpha

and

\beta

are constants

x

and

y

are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection

\theta

and the maximum height attained

H

are respectively given by

View answer and explanation

Correct option: A

{\tan ^{ - 1}}\alpha ,\,\,\frac{{{\alpha ^2}}}{{4\beta }}

Option A (

{\tan ^{ - 1}}\alpha ,\,\frac{{{\alpha ^2}}}{{4\beta }}

) is correct because it matches the angle of projection and maximum height derived from the trajectory equation.

More Questions from MOTION IN PLANE

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\mathrm{\vec P} = \mathrm{\vec A} + \mathrm{\vec B}\,\,\mathrm{and}\,P = A + B.

\mathrm{\vec A}\,and\,\mathrm{\vec B}

0^{\circ}

90^{\circ}

\mathrm{If}\,(\mathbf{a}+3 \mathbf{b}) \cdot(7 \mathbf{a}-5 \mathbf{b})=0

(\mathbf{a}-4 \mathbf{b}) \cdot(7 \mathbf{a}-2 \mathbf{b})=0

\mathbf{a}

\mathbf{b}

\theta

\left(90^{\circ}-\theta\right)

39.2 \mathrm{~ms}^{-1}

50 \mathrm{~m}

{g}=9.8 \mathrm{~ms}^{-2}