Kinematics
![Picture](/uploads/1/5/1/5/15153444/8036407.jpg?145)
Kinematics is study of motion and development of equations that explains the movements of objects. Kinematics can occur in one, two, and three-dimensional spaces. Kinematics allows us to find out many interesting facts in the game of American football such as the maximum height that a football reaches when being thrown. In this section we will be looking at the potential distance, max height, and flight time that a football can achieve while in motion. Michael Vick throws a pass to a receiver down field with a resultant velocity of 30m/s [30 degrees above the horizontal] at a height of 1.4m. We will ignore air resistance. The receiver goes down the field and completes the catch for a gain of 25 yards during the flight the ball follows a parabolic path. In order to determine the distance travelled, max height, and time of flight we must find the initial vertical and horizontal velocity of the ball using SOHCAHTOA.
The initial resultant velocity of the ball is 30m/s [30 degrees above the horizontal] VRi = 30m/s
The initial vertical velocity of the ball is = (30m/s)(sin30) Viy(velocity vertical initial) = 15m/s [up].
The initial horizontal velocity of the ball is = (30m/s)(Cos30) Vix(velocity horizontal initial) = 25.99m/s [forwards]
The initial resultant velocity of the ball is 30m/s [30 degrees above the horizontal] VRi = 30m/s
The initial vertical velocity of the ball is = (30m/s)(sin30) Viy(velocity vertical initial) = 15m/s [up].
The initial horizontal velocity of the ball is = (30m/s)(Cos30) Vix(velocity horizontal initial) = 25.99m/s [forwards]
Time of flight and distance travelled
To find time of flight we plug in all the known components we have.
Given: Viy = 15m/s [up] Vfy = 0m/s Acceleration due to gravity = 9.8m/s^2[down]
Required: time
Analysis: Vfy = t*a + Viy
Substitute: 0 = T*(9.8m/s) + 15m/s[up]
= 15m/s[up] / 9.8m/s
Solution: T = 1.53 x 2 because this is only time to max height
Therefore the time of flight is 3.06 seconds.
We can now use our horizontal components to find time to solve for distance travelled.
Given: Vix = 25.99m/s [Forward] Vfx = 25.99m/s Acceleration due to gravity = -9.8m/s^2 time = 1.53 seconds
Required: Distance (d)
Analysis: d = 1/2t (vi + vf)
Substitute: d = 1/2(3.06) (25.99m/s + 25.99m/s)
Solution: d = 79.53m
Therefore the ball travells 79.53m.
Given: Viy = 15m/s [up] Vfy = 0m/s Acceleration due to gravity = 9.8m/s^2[down]
Required: time
Analysis: Vfy = t*a + Viy
Substitute: 0 = T*(9.8m/s) + 15m/s[up]
= 15m/s[up] / 9.8m/s
Solution: T = 1.53 x 2 because this is only time to max height
Therefore the time of flight is 3.06 seconds.
We can now use our horizontal components to find time to solve for distance travelled.
Given: Vix = 25.99m/s [Forward] Vfx = 25.99m/s Acceleration due to gravity = -9.8m/s^2 time = 1.53 seconds
Required: Distance (d)
Analysis: d = 1/2t (vi + vf)
Substitute: d = 1/2(3.06) (25.99m/s + 25.99m/s)
Solution: d = 79.53m
Therefore the ball travells 79.53m.
max height
![Picture](/uploads/1/5/1/5/15153444/2649909.jpg?178)
To find the max height of this ball we must use vertical components.
Given: Viy = 15m/s [up] Vfy = 0m/s[up] at max height Acceleration due to gravity = -9.8m/s^2
Required: Distance (d)
Analysis: Vfy^2 = Viy^2 + 2a(d)
Substitute: 0 = (15m/s)^2 + 2(*9.8m/s^2)d
Solution: -225m/s = (-19.6m/s^2) d
-225m/s / (-9.6m/s^2)= d
11.48m + 1.4 m= d
You have to factor in the height that the ball is being thrown at so you at 1.4m. Therefore the max height of the ball is 12.88m.
Given: Viy = 15m/s [up] Vfy = 0m/s[up] at max height Acceleration due to gravity = -9.8m/s^2
Required: Distance (d)
Analysis: Vfy^2 = Viy^2 + 2a(d)
Substitute: 0 = (15m/s)^2 + 2(*9.8m/s^2)d
Solution: -225m/s = (-19.6m/s^2) d
-225m/s / (-9.6m/s^2)= d
11.48m + 1.4 m= d
You have to factor in the height that the ball is being thrown at so you at 1.4m. Therefore the max height of the ball is 12.88m.
POsition-time graph
![Picture](/uploads/1/5/1/5/15153444/3615685.jpg?410)
The graph to the right represents the parabolic journey of an NFL football being thrown up in the air (vertical position). As the player releases the football it accelerates in and horizontal and vertical direction until it reaches a max height. As the ball reaches max height it then begins to descend towards the ground. The graph would differ if it was representing a horizontal distance travelled as the line would go straight up diagonally and would not go back to zero because it would keep travelling away from the thrower.
Velocity-time graph
![Picture](/uploads/1/5/1/5/15153444/2922873.jpg?434)
The velocity-time graphs shows the change in velocity of the football during the flight of the football. The ball begins at a velocity of about 29m/s as the ball is released from the players hands. As the ball beings to fly through the air in a enclosed stadium the ball will encounter air resistance which means that the velocity of the ball will begin to decrease. Even though the velocity of the ball is decreasing the ball is still rising in the air.When the ball reaches max height the velocity becomes zero. Once the ball is finished travelling at max height the ball begins to descend with a negative velocity.