Parametric Problem: Kicking a football (Getting Ready For Test)

Question

As I told you on title I'm getting ready for a test. I have this Test-Review Problem....

A football is kicked from the ground with an initial velocity of $28$ ft/sec at an angle of $28^\circ$. How far does the ball travel? Assuming the football is kicked at the $0$ yard line of the home team, where does the football land? What was the maximum height of the football and when did it occur? SHOW ALL WORK

This question was on my test review for one of the last few problems. I have no idea how to do this, and this is probably gonna be worth a lot of points on my test. So please help me out guys, you guys are my only hope in math!

Answer 1

You break the motion of the football into its horizontal and vertical components. (x,y)

the initial velocity:

$v_x(0) = 28 \cos 28 ft/s\\ v_y(0) = 28 \sin 28) ft/s$

Gravity is a constant force pulling the ball back to the ground.

Without calculus, best I can say is that there is a standard formula for an object falling under gravity.

$y = h(0) + v_y(0) t - 16 t^2$ (in feet)

If you are in metric $y = h(0) + v_y(0) t - 4.9 t^2$ (meters)

$y(t) = 0 + (28 \sin 28) t -16 t^2\\ x(t) = (28 \cos 28) t $

The ball hits the ground when $y(t) = 0$

$y = 28 \sin 28 t - 16 t^2 = 0$ solve for $t$.

When you know $t,$ plug that in to $x(t)$ and you will have the distance of your kick.

maximum height $y (t)$ is a parabola. To find the vertex of a parabola. Complete the square... $y = a(t-h)^2 + v$

And find v. Since the ball starts and ends at the same height. The ball reaches the vertex in exactly half the time of the total flight of the ball.

Where does the ball land? The foot ball field is measured in yards. Convert from feet to yards. If the kick is greater than 50 yards, then you count backward after crossing the 50 yard line.

Answer 2

Here is a brief review of the relevant 2D kinematics. I take the origin to be the initial location of the ball, the horizontal $x$-axis to be parallel to the ground, and the vertical $y$-axis to be perpendicular to the ground; they are oriented so that the initial velocity components are both positive.

We first consider the horizontal motion of the ball. There are no forces acting in the $x$-direction---air resistance is ignored, for example---and therefore the ball has no acceleration in the $x$-direction. But the acceleration is defined to be the rate at which the velocity is changing, so we conclude that the $x$-component of the velocity vector $\vec{v}$ doesn't change: $\boxed{{v_x(t)=v_{x0}}}$ where $v_{x0}$ is the $x$-component of the initial velocity $\vec{v}_0$.

Since this velocity is constant, the change in position along the $x$-axis is simply the velocity times time. Consequently $\boxed{x(t)=x_0+v_{x0}t}$ where $x_0$ is the initial $x$-component of the position vector $\vec{r}$.

We now consider the vertical motion. If there were no forces acting in the vertical direction, there would be no acceleration in that direction and therefore the same remarks as above would apply: $v_y(t)=v_{y0}$, $y(t)=y_0+v_{y0} t$. But the gravitional force acts to pull the ball downwards, giving a constant acceleration $a_y=-g=-32\text{ ft/s}^2$. Since the acceleration is the rate at which the velocity changes, this modifies the velocity result to $\boxed{v_y(t)=v_{y0}-g t}$. (Compare with the earlier result for $x(t)$.)

This leaves $y(t)$. For this, precalculus isn't really enough since we need to know how the position changes when the velocity is changing. I will bypass this question, and simply assert that one obtains $\boxed{y(t)=y_0+v_{y0}t-\frac12 gt^2}$.

The boxed equations correspond to the kinematic equations for the motion. (Other combinations can be obtained algebraically from these, e.g. $y$ as a function of $v_{y0}$). These should be the starting point for your solution; the remaining task is to obtain the initial values $x_0,y_0,v_{x0},v_{y0}$ and determine equations that can be solved for the desired quantities. I leave this to the reader.