JEE PHYSICS MCQ: MOTION IN PLANE

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For angle

\ldots X \ldots

the projectile has maximum range and it is equal to ...X.... Here,

\mathrm{X}

and

\mathrm{Y}

refer to

View answer and explanation

Correct option: C

\frac{\pi}{4}

and

\frac{v_0^2}{g}

Option C (

\frac{\pi}{4}

and

\frac{v_0^2}{g}

) is correct because the maximum range of a projectile is achieved at an angle of 45 degrees (

\frac{\pi}{4}

radians) and is equal to

\frac{v_0^2}{g}

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}