If the velocity of a long jumper during take-offat an angle of 30 degrees is 42 f/s, how fast is he moving forward (Vx) and how fast is he moving upward (Vy) ? The answers should be in international units (m/s). 1 foot = 0.3048 m

Vx = 11.0865 (m/s)

Vy = 6.4008 (m/s)

Explanation:

Taking into account that 1m is equal to 0.3048 ft, the takeoff speed in m / s will be:

V = 42 (ft/s) * 0.3048 (m/ft) = 12.8016 (m/s)

The take-off angle is equal to 30 °, taking into account the Pythagorean theorem the velocity on the X axis will be:

Vx = 12.8016 (m/s) * cos (30°) = 11.0865 (m/s)

And for the same theorem the speed on the Y axis will be:

Vy = 12.8016 (m/s) * sen (30°) = 6.4008 (m/s)