NEET UG PHYSICS MCQ: MOTION IN PLANE

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Application +4 | -1

For a particle projected with speed of 20 m/s from ground, the kinetic energy at top most point of trajectory is

\left( {\frac{1}{4}} \right)th

of the initial kinetic energy. The angle of projection from horizontal is

View answer and explanation

Correct option: B

60^\circ

Option B (

60^\circ

) is correct. The angle of projection is determined by the ratio of kinetic energies at the initial point and the top of the trajectory.

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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)