NEET UG PHYSICS MCQ: MOTION IN PLANE

1 Easy

1 of 1

Application +4 | -1

A particle is projected at an angle of

{60^0}

above the horizontal with a speed of

10\mathrm{\;m}{\mathrm{s}^{ - 1}}

. After some time the direction of its velocity makes an angle of

{30^0}

above the horizontal. The speed of the particle at this instant is

View answer and explanation

Correct option: D

\frac{{10}}{{\sqrt 3 }}\mathrm{m}{\mathrm{s}^{ - 1}}

Option D (

\frac{{10}}{{\sqrt 3 }}\mathrm{\;m}{\mathrm{s}^{ - 1}}

) is correct because the horizontal component of velocity remains constant, allowing us to solve for the final speed.

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\mathrm{\vec a}\,and\,\mathrm{\vec b}

\mathrm{\vec a}\,and\,\mathrm{\vec b}

\overrightarrow P \,\,.\,\,\overrightarrow Q = 0\,and\,\overrightarrow P \,\, \times \,\,\overrightarrow Q = 0

\overrightarrow Q \,and\,\overrightarrow R

\vec{A}=3 \hat{i}+4 \hat{j}

\vec{B}=6 \hat{i}+8 \hat{j}

A

B

\overrightarrow{\mathrm{A}}

\overrightarrow{\mathrm{B}}

{\cot ^{ - 1}}\left( 2 \right)