JEE PHYSICS MCQ: MOTION IN PLANE

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Application +4 | -1

A particle is projected with an initial velocity

\vec{v}=(2 \hat{i}+4 \hat{j}) \mathrm{m} / \mathrm{s}

. The equation of trajectory of the particle will be

\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)

(Assume horizontal direction as

x

-axis and vertically upward as positive y-axis)

View answer and explanation

Correct option: A

y=2 x-\frac{5 x^2}{4}

Option A (

y=2 x-\frac{5 x^2}{4}

) is correct because the equation of the trajectory is derived by eliminating time from the x and y components of motion.

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}