JEE PHYSICS MCQ: MOTION IN PLANE

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Understanding +4 | -1

A bullet is fired with a velocity u making an angle of

{60^0}

with the horizontal plane. The horizontal component of the velocity of the bullet when it reaches the maximum height is

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Correct option: D

u/2

Option D (u/2) is correct because the horizontal component of velocity remains constant throughout projectile motion and is equal to

u \cos{\theta}

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}