Quadratic functions: [Diagram Included] Football game scenario

[Swan]

Homework Statement

You are the quarterback for the Quinte Saints Football team. You are in the middle of the COSSA gold medal game and you see your receiver is wide open down the field beside the sideline. If he catches the ball, you win the game. However, the biggest guy Joey from the opposing team who is 190 cm tall is running towards you. You decide to throw the ball so that the highest part of the path is 1.5 m over Joey's head to avoid him reaching it if he jumps. You throw the ball at your receiver releasing the football at your head level when Joey is 5 m away from you.

The goal of this task is to figure out if your teammate can catch the ball and win the game.

Assumption variable: Your Height (to the nearest centimeter): 150 cm. **NOTE**: We had to assume what the quarterback's height was and in this my height is 150 cm.

a) Draw a sketch of the situation including the path of the ball (assume no wind). Fill in all information you know at this time.​

b) Determine the equation representing the path of the football.​

c) Your receivers typically catch the ball 1 m from the ground. If your player is 9.5 m away, and can run within 2 m of his initial location to catch the ball, does he catch the ball to win the game? Justify.​

Homework Equations

None

The Attempt at a Solution

a)

b) I have no idea how to go about this.​

c) I say he would catch the ball to win the game because the receiver can cover 2 m of horizontal distance if needed.​

 

[tms]

a) The diagram looks okay except that you left out the details of the receiver: the height and radius in which he can catch the ball.

b) Start by writing down the kinematics equations. Since you are not given Joey's speed, I think you can assume it is enough slower than the ball's speed that it can be ignored. You don't know the speed at which the ball is thrown, or the angle, but you do know where it has to be to avoid Joey, and approximately where it has to land to be caught. With that information it should be possible to find the solution.

c) You can't answer (c) until you find the equation in (b). It is possible that the ball speed needed to avoid Joey will be such that the ball won't land close enough to the receiver.

 

[Swan]

no information about the receiver is given and also no speed is given throughout this question

 

[tms]

You are told that the ball will be caught 1 m above the ground within 2 m of the receiver's initial position. Those conditions put some constraints on the initial velocity and angle of the ball. Without those constraints, if all you had to do was avoid the defender, you'd just have to use a very high speed and a high angle.

 

[Nickie]

By the time the ball reaches 9.5 m, it would have already reached its highest point and started to descend, giving the receiver enough time to cover the remaining 2 m and catch the ball at 1 m above the ground.


I would like to provide a response to this scenario by analyzing the situation using mathematical equations and principles.

Firstly, we can represent the path of the football using a quadratic function, where the height of the ball (y) is a function of the horizontal distance (x) it has traveled. The general form of a quadratic function is y = ax^2 + bx + c, where a is the coefficient of the squared term, b is the coefficient of the linear term, and c is the constant term.

In this scenario, we can assume that the ball is thrown with an initial velocity of v0 at an angle of 45 degrees (optimal angle for maximum distance) and there is no air resistance. Therefore, the equation for the path of the ball can be written as y = -9.8x^2/(2v0^2cos^2θ) + xtanθ + h, where θ is the angle of release, h is the initial height of the ball (in this case, the height of the quarterback's head), and v0 is the initial velocity.

Using the given information, we can calculate the value of v0 by using the distance formula, d = v0t + 1/2at^2, where d is the horizontal distance traveled (9.5 m), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time taken for the ball to reach 9.5 m. By rearranging the equation, we get t = (2d/a)^0.5 = 1.38 seconds.

Next, we can calculate the angle of release, θ, using the equation tanθ = v0y/v0x, where v0y is the vertical component of the initial velocity (v0sinθ) and v0x is the horizontal component (v0cosθ). Plugging in the values, we get θ = 45 degrees.

Now, we can plug in these values in the equation for the path of the ball and solve for the height of the ball at a horizontal distance of 5 m (when Joey is 5 m away). This will