A football is kicked 66.6 meters. If it is in the air for 5.39 seconds, with what initial velocity was it kicked?
magnitude and direction above horizontal
NEED HELP BAD
IT IS NOT COMING OUT RIGHT
To find the initial velocity with which the football was kicked, we can use the equations of motion for projectile motion. Specifically, we can use the equation of motion that links the vertical displacement, initial vertical velocity, time, and acceleration due to gravity:
y = v₀y * t - (1/2) * g * t²,
where:
- y is the vertical displacement (in this case, it is the height the ball reaches, which is 0),
- v₀y is the initial vertical velocity (which we want to find),
- t is the time (5.39 seconds), and
- g is the acceleration due to gravity (approximately 9.8 m/s²).
Since the initial vertical velocity (v₀y) is zero at the highest point of a projectile's trajectory, we can simplify the equation to:
0 = - (1/2) * g * t².
Now, we can solve for the acceleration due to gravity (g):
g = 2 * 0 / (t²),
g = 0.
Since g is equal to zero, it means that the football was kicked horizontally. Therefore, the initial vertical velocity (v₀y) is zero.
To find the initial horizontal velocity (v₀x), we can use the equation that relates the horizontal displacement, initial horizontal velocity, and time:
x = v₀x * t.
Rearranging the equation to solve for v₀x:
v₀x = x / t.
In this case, the horizontal displacement (x) is equal to 66.6 meters, and the time (t) is 5.39 seconds.
v₀x = 66.6 m / 5.39 s,
v₀x ≈ 12.34 m/s.
Therefore, the initial velocity with which the football was kicked is approximately 12.34 m/s. Since we know that the initial vertical velocity is zero, the initial velocity is entirely horizontal.
So, the magnitude of the initial velocity is 12.34 m/s, and the direction is horizontally to the right.
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°=...