## Monday, February 8, 2021

### PHYSICS IN FOOTBALL - KINEMATICS

Kinematics is the mechanics of the motion of objects, disregarding the forces that cause that motion. So how can we apply this to football?
We will: 1. Calculate the maximum speed the football will reach when it is thrown
2. Calculate how much the ball accelerates when thrown
3. Calculate how far the ball will be thrown given the velocity, launch height, and launch angle

1. Maximum Speed
Standing 10.0m away from a wall, I threw a football at the wall with a time of 1.04s from the balls release to the ball hitting the wall. I can figure out the maximum speed the ball reaches when I threw it by using the variables initial speed (0m/s since the ball wasn’t moving before thrown), distance travelled (10.0m),time taken (1.04s), and final velocity (v2).
d=0.5(v1+v2)t
10.0m=0.5(0m/s+v2)(1.04s)
10.0m/[(0.5)(1.04s)]=(0.5)(v2)(1.04s)/[(0.5)(1.04s)]
10.0m/0.52s=v2
v2=19.230769
Therefore the maximum speed the ball reached was 19.2m/s

2. Acceleration
Using the velocity we found, we can use it along with the initial speed(0m/s since the ball wasn't moving before it was thrown), and the time taken (1.04s) to figure out the acceleration of the ball.
a=(V2-V1)/t
a=(19.2m/s-0m/s)/1.04s
a=19.2m/s/1.04s
a=18.46153846
Therefore the acceleration of the football when I threw it was 18.5m/s^2 3. Projectile Motion
Using the maximum speed we have got earlier (19.2m/s), the height of the ball when it was thrown from my hand (1.5m), and the launch angle (25 degrees) we can figure out how far the football will travel.
v1v=19.2m/s(sin25)
vh=19.2m/s(cos25)
Horizontal Variables(Right=+): vh=19.2m/s(cos25), dh=want, t=?
Vertical Variables(up=+): v1v=19.2m/s(sin25), a=-9.8m/s, dv=-1.5m, t=?

dv=(v1)(t)+(0.5)(a)(t2)
-1.5m=[19.2m/s(cos25)](t)+(0.5)(9.8m/s)(t)^2
-1.5m=[19.2m/s(cos25)](t)+(4.9m/s)(t)^2
*Let t=x
4.9m/s x^2-[19.2m/s(cos25)]x-1.5m t=3.961177404s
Now that we know the time taken for the ball to reach the ground from the release, we can figure out how far it has travelled using the time taken and horizontal velocity.
vh=dh/t
dh=(vh)(t)
dh=[19.2m/s(cos25)](3.961177404s)
dh=(19.03109399m/s)(3.961177404s)
dh=75.38553948m
Therefore, the football will travel 75. 4m in total