Question

. An aeroplane takes off at an angle of 60o to the horizontal. If the component of the velocity along the horizontal is 200m/s, what is the actual velocity? Also find the vertical component of its velocity.

Answer 1

The horizontal velocity

v_x=v cos⁡θ

So

v=v_x/cos⁡θ =(200 m/s)/cos⁡〖60°〗 =400 m/s

The vertical velocity

v_y=v sin⁡〖θ=400 m/s×sin⁡〖60°=364 m/s〗 〗