A Football is kicked 66.6 meters in 5.39 seconds. What was the initial velocity it was kicked?

magnitude and direction above horizontal

x = vx*t

y = vy*t - 0.5*9.8*t^2

Assuming it hits the ground at 5.39 seconds:

66 = vx*5.39
0 = vy * 5.39 - 0.5*9.8*5.39^2

Where vx is the speed in the x direction, vy is the speed in the y direction.

The magnitude of the speed is

(vx^2 + vy^2)^0.5

The direction of the speed is given by

tan(x) = vy / vx
Solve for the angle x

how do you get the vy by it self when trying to get tionhe speed in the y direct

To determine the initial velocity at which the football was kicked, you can use the following equation of motion:

\( V = \frac {d}{t} \)

Where:
V = Initial velocity
d = Distance traveled
t = Time taken

In this case, the distance traveled (d) is given as 66.6 meters, and the time taken (t) is given as 5.39 seconds. By substituting these values into the equation, we can find the initial velocity:

\( V = \frac {66.6 \, \text{m}}{5.39 \, \text{s}} \)

Calculating this equation will give us the magnitude of the initial velocity. To find the direction above horizontal, we need the vertical component of the velocity. We can assume the football was kicked horizontally, so the vertical component of the initial velocity would be zero.

Therefore, the magnitude of the initial velocity is:

\( V = \frac {66.6 \, \text{m}}{5.39 \, \text{s}} \, = \, \text{12.34 m/s} \)

The direction is horizontal.