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At takeoff, the horizontal and vertical velocity of long jumper are 57.8 m/s and 5.6 m/s respectively. What is the resultant velocity

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Apply the Pythagorean theorem to get the resultant velocity:

V =
\sqrt{Vx^(2)+Vy^(2)}

Given values:

Vx = 57.8m/s

Vy = 5.6m/s

Plug in and solve for V:

V = 58.1m/s

EDIT: Let's get the direction of the resultant velocity as well.

This equation will give the angle of the velocity as measured off of the ground:

θ = tan⁻¹(Vy/Vx)

Again, the given values are:

Vx = 57.8m/s

Vy = 5.6m/s

Plug in the values and solve for the angle θ:

θ = tan⁻¹(5.6/57.8)

θ = 5.5°

The resultant velocity is oriented 5.5° off the ground.

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