A projectile is fired at an angle of 30° to the horizontal such that the vertical component of its initial velocity is $80{\rm{\;m}}{{\rm{s}}^{ - 1}}$. Its time of flight is T. Its velocity at $t = \frac{T}{4}$ has a magnitude of nearly

$140\,{\rm{m}}{{\rm{s}}^{ - 1}}$

$200{\rm{\;m}}{{\rm{s}}^{ - 1}}$

$300{\rm{\;m}}{{\rm{s}}^{ - 1}}$

$100\,{\rm{m}}{{\rm{s}}^{ - 1}}$

JEE PHYSICS MCQ: MOTION IN PLANE

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Application +4 | -1

A projectile is fired at an angle of 30° to the horizontal such that the vertical component of its initial velocity is

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

. Its time of flight is T. Its velocity at

t = \frac{T}{4}

has a magnitude of nearly

View answer and explanation

Correct option: C

140\,\mathrm{m}{\mathrm{s}^{ - 1}}

Option C (

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

) is correct because it is the magnitude of the velocity at t=T/4, calculated using projectile motion principles.

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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}