A football is place kicked with a velocity having a vertical component of 12 m/s and a horizontal component of 6 m/s. Find the resultant velocity. What is the angle of release with respect to the horizontal?

The velocity is given by:

V = √ (Vx²+Vy²)

V = velocity, Vx = horizontal velocity, Vy = vertical velocity

Given values:

Vx = 6m/s, Vy = 12m/s

Plug in and solve for V:

V = √ (6²+12²)

V = 13.42m/s

Now find the direction:

θ = tan⁻¹ (Vy/Vx)

θ = angle of velocity off horizontal, Vy = vertical velocity, Vx = horizontal velocity

Given values:

Vx = 6m/s, Vy = 12m/s

Plug in and solve for θ:

θ = tan⁻¹ (12/6)

θ = 63.4°

The resultant velocity is 13.42m/s at an angle of 63.4° off the horizontal.