A football is kicked 66.6 meters. If the ball is in the air for 5.39 seconds, with what initial velocity was it kicked?

NEED HElP

L=vₒ²•sin2α/g = 66.6

t= 2vₒ•sinα/g = 5.39
vₒ = t•g/2•sinα
L=vₒ²•sin2α/g= t²•g²•2•sinα•cosα/g•4•sin²α= t²•g/2•tanα
tanα = t²•g/2•L
α= arctan t²•g/2•L = arctan (5.39²•9.8/2•66.6)=64.9°.
vₒ = t•g/2•sinα=5.39•9.8/2•sin 64.9°=...

To find the initial velocity, we can use the formula for the vertical displacement of an object in freefall:

h = (vi * t) + (1/2 * g * t^2)

Where:
h = vertical displacement (which is the height of the ball, equal to 66.6 meters)
vi = initial velocity
t = time in the air (which is 5.39 seconds)
g = acceleration due to gravity (approximately 9.8 m/s^2)

Rearranging the formula to solve for the initial velocity (vi), we have:

vi = (h - (1/2 * g * t^2)) / t

Let's substitute the given values into the formula and calculate the initial velocity:

vi = (66.6 - (1/2 * 9.8 * 5.39^2)) / 5.39

Calculating further:

vi = (66.6 - (1/2 * 9.8 * 29.0921)) / 5.39
= (66.6 - (143.3929)) / 5.39
= (66.6 - 143.3929) / 5.39
= (-76.7929) / 5.39
= -14.250 m/s

Therefore, the initial velocity at which the ball was kicked is approximately -14.250 m/s. Note that the negative sign indicates that the initial velocity is in the opposite direction of gravity.

To find the initial velocity of the football, we can use the kinematic equation:

s = ut + (1/2)at^2

where:
s = displacement (distance travelled) = 66.6 meters
u = initial velocity
t = time = 5.39 seconds
a = acceleration (assumed to be in this case as 0, as there is no mention of acceleration)

Rearranging the equation to solve for initial velocity (u), we have:

u = (s - (1/2)at^2) / t

Since we assume there is no acceleration, the equation simplifies to:

u = s / t

Substituting the given values:

u = 66.6 meters / 5.39 seconds
u ≈ 12.34 meters per second

Therefore, the initial velocity of the football was approximately 12.34 meters per second.