Question

An aeroplane takes off at an angle of 30 degree to the horizontal. if the component of its velocity along teh horizontal is 250 km/hr, what is the actual velocity. find also its vertical component

Answer 1

let the actual velocity be v

v along x=v cos(theta)
250 = v cos 30
250=(v × 3^1/2)÷2
500÷(3^1/2) =v

v=500/underroot3 km/hr

v along vertical dir. = v sin 30
= v/2
=500/(2×root3) km/hr

Answer 2

$\huge {\bf {\underline {Question}}}$

An aeroplane takes off at an angle of 30 degree to the horizontal. if the component of its velocity along teh horizontal is 250 km/hr, what is the actual velocity. find also its vertical component.

$\huge {\bf {\underline {Answer}}}$

Let the velocity of the plane be V m/s.

Horizontal component of velocity =

$V cos 30° = 250 km/h \\ \implies V \frac {\sqrt {3}}{2} = 250 \\ \implies V = \frac {250 \times 2}{\sqrt {3}} \\ \implies V = \frac {500}{\sqrt {3}} \\ \implies V = \frac {500}{1.732} \\ \implies V = 288.7 \: km/h$

Vertical component of velocity =

$V sin 30° = \frac {V}{2} = \frac {288.7}{2} = 144.35\:km/h$

Hope it helps dear....