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3.2 , motion in plane — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysics3.2 , motion in plane1 Mark
MCQ
A particle is projected from the ground with an initial speed $\upsilon $ at an angle $\theta $ with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is
  • A
    $\frac{\upsilon }{2}\sqrt {1 + 2\,\,{{\cos }^{2\,}}\theta } $
  • B
    $\frac{\upsilon }{2}\sqrt {1 + {{\cos }^{2\,}}\theta } $
  • ✓
    $\frac{\upsilon }{2}\sqrt {1 + 3\,\,{{\cos }^{2\,}}\theta } $
  • D
    $\upsilon \,\cos \,\theta $
✓

Answer

Correct option: C.
$\frac{\upsilon }{2}\sqrt {1 + 3\,\,{{\cos }^{2\,}}\theta } $
c
Average velocity $=\frac{\text { Displacement }}{\text { Time }}$

$v_{w_{w}}=\frac{\sqrt{H^{2}+\frac{R^{2}}{4}}}{T / 2}$

Here, $\quad \mathrm{H}=$ maximum height $=\frac{v^{2} \sin ^{2} \theta}{2 g}$

$\mathrm{R}=\mathrm{range}=\frac{\mathrm{v}^{2} \sin 2 \theta}{\mathrm{g}}$

and $\quad \mathrm{T}=$ time of flight $=\frac{2v \sin \theta}{\mathrm{g}}$

$\therefore \quad \mathrm{U}_{\mathrm{av} .}=\frac{\mathrm{v}}{2} \sqrt{1+3 \cos ^{2} \theta}$

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      Column $-I$

    Angle of projection

     Column $-II$
  $A.$ $\theta \, = \,{45^o}$   $1.$ $\frac{{{K_h}}}{{{K_i}}} = \frac{1}{4}$
  $B.$ $\theta \, = \,{60^o}$   $2.$ $\frac{{g{T^2}}}{R} = 8$
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  $D.$ $\theta \, = \,{\tan ^{ - 1}}\,4$   $4.$ $\frac{R}{H} = 4$

$K_i :$ initial kinetic energy

$K_h :$ kinetic energy at the highest point

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