JEE PHYSICS MCQ: MOTION IN PLANE

1 Easy

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Analyzing +4 | -1

Two particles are projected from the same point with the same speed, at different angles

\theta_1

and

\theta_2

to the horizontal. They have the same horizontal range. Their times of flight are

t_1

and

t_2

respectively incorrect statement is.

View answer and explanation

Correct option: C

\frac{t_1}{t_2}=\tan \theta_2

Option C (

\frac{t_1}{t_2}=\tan \theta_2

) is correct because the correct relationship is

\frac{t_1}{t_2} = \frac{\sin \theta_1}{\sin \theta_2}

, and since

\theta_2 = 90 - \theta_1

\frac{t_1}{t_2} = \tan \theta_1

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}