Question

An aeroplane takes off at an angle of ${60}^o$ to the horizontal. If the velocity of the plane is $200km{h}^{-1}$, calculate its horizontal and vertical component of its velocity.

• A. ${\nu }_y\sin \theta =300\sqrt{3}km{h}^{-1}$
• B. ${\nu }_y\sin \theta =100\sqrt{3}km{h}^{-1}$
• C. ${\nu }_y\sin \theta =200\sqrt{3}km{h}^{-1}$
• D. ${\nu }_y\sin \theta =400\sqrt{3}km{h}^{-1}$

Answer 1

Answer: (B)

${\nu }_y\sin \theta =100\sqrt{3}km{h}^{-1}$

Here $\nu =200km{h}^{-1}$

$\theta ={60}^o$
$\therefore$ Horizontal component
${\nu }_x=\nu \cos \theta =200\cos {60}^o$
$=200\times \frac{1}{2}=100km{h}^{-1}$
Vertical component
${\nu }_y\sin \theta =200\sin {60}^o=200\times \frac{\sqrt{3}}{2}=100\sqrt{3}km{h}^{-1}$