Projection angle

[AusCharger]

Could somebody please tell me how to calculate the angle of projection using a Horizontal and Vertical velocity? By the way I did run a search first and found diddley.

Cheers

[James R]

tan(angle) = vertical velocity / horizontal velocity.

[AusCharger]

Is that all? Oops, feel like bit of an idiot. Thanks a lot for that James.

[James R]

Velocity is a vector, so if the magnitude is v and the angle it t (which I'll use instead of theta), then the components of the velocity are:

vx = v cos t
vy = v sin t

Draw a right-angled triangle with angle t between the x direction and the hypotenuse. Then

sin t = opposite / hypotenuse = vy/v ... (1)
cos t = adjacent / hypotenuse = vx/v ... (2)

which leads to the above relationships.

If we divide equation (1) by (2), we get:

sin t / cos t = (vy/v) / (vx/v) = vy/vx = tan t.

Also:

v^2 = vx^2 + vy^2

using Pythagoras.