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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 projectile is fired with a speed $u$ at an angle $\theta$ with the horizontal. Its speed when its direction of motion makes an angle ‘$\alpha $’ with the horizontal is
  • A
    $u\,\, sec\theta \,\,cos\alpha$
  • B
    $u\,\, sec\theta\,\, sin\alpha$
  • ✓
    $u \,\,cos\theta \,\,sec\alpha$
  • D
    $u\,\, sin \theta \,\,sec\alpha$
✓

Answer

Correct option: C.
$u \,\,cos\theta \,\,sec\alpha$
c
Horizontal component of velocity $=v_{x}=u \cos \theta$

 vertical component of velocity $v_{y}=u \sin \theta-g t$

angle of velocity with horizontal $=\alpha$

$\tan a=\frac{v_{y}}{v_{x}}=\frac{(u \sin \theta-g t)}{(4 \cos \theta)}$

$t=\frac{u(\sin \theta-\cos \theta \tan \alpha)}{g}$

$v_{y}=u \cos \theta \tan \alpha$

$v_{x}=u \cos \theta$

So the speed of the projectile $=\sqrt{\left(\left(v_{x}\right)^{2}+\left(v_{y}\right)^{2}\right)}$

 $=u \cos \theta \sec \alpha$

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