Simple motion question

Question

The question states that an airplane takes off at v = 67m/s at an angle of 45º. To calculate the "magnitude of the vertical component of the velocity", can I use "67" as hypotenuse on a right-angle triangle diagram?

Thanks!

Answer 1

Sturm's answer is right, and yes, 67 is the hypotenuse.  This gives you: sin(45) = y / 67, which simplifies to: 67sin(45) = y

Answer 2

The vertical component is the y-component. With that being said, use the product of the velocity and the sine of the angle.

v(y) = v sin θ = 67 m/s x sin(45º) = 47 m/s