Q58P - University Physics with Modern Physics

Question

In Canadian football, after a touchdown the team has the opportunity to earn one more point by kicking the ball over the bar between the goal posts. The bar is 10.0 ft above the ground, and the ball is kicked from ground level, 36.0 ft horizontally from the bar (Fig. P3.58). Football regulations are stated in English units, but convert them to SI units for this problem. (a) There is a minimum angle above the ground such that if the ball is launched below this angle, it can never clear the bar, no matter how fast it is kicked. What is this angle? (b) If the ball is kicked at $\mathbf{45}\mathbf{.}\mathbf{0}\mathbf{°}$ above the horizontal, what must its initial speed be if it is just to clear the bar? Express your answer in m/s and in km/h .

                           

Step-by-Step Solution

Answer

a) The minimum angle above the ground is $15.5°$ . 

b) The initial speed of the ball is $43.9 km/h$ . 

Step 1: Introduction

The second law of motion is given by,

 

$s=ut+\frac{1}{2}a{t}^2$

 

Here s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Step 2: Given data.

The height of the bar is 10.0 ft above the ground.

Horizontal distance of the ball is 36.0 ft from the bar.

The ball is kicked at an angle of $45°$ above the horizontal.      

Step 3: Calculate the minimum angle above the ground.

(a)

 

The minimum angle is given by,

$$\begin{gathered} \tan \theta =\left(\frac{y}{x}\right) \\ \theta ={\tan }^{-1}\left(\frac{y}{x}\right) \end{gathered}$$ 

Substitute 10.0 ft for y and 36.0 ft for x in the above equation.

 $$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{10.0}{36.0}\right) \\ =15.52° \end{gathered}$$

Therefore the minimum angle above the ground is $15.52°$ .

Step 4: Calculate the minimum speed of the ball.

(b)

 

Converts the units of horizontal distance into meter,

 

$$\begin{gathered} x=36.0\left(0.3048\right) \\ =10.9782m \end{gathered}$$ 

 

Converts the units of vertical distance into meter,

 

$$\begin{gathered} y=10.0\left(0.3048\right) \\ =3.048m \end{gathered}$$ 

 

As in horizontal there will be no acceleration due to gravity.

 

Using the kinematic equation in horizontal direction,

$$\begin{gathered} x=v_{x}t \\ =v_0\cos 45° \end{gathered}$$ 

Here $v_0$ is the initial velocity, t is the time.

$t=\frac{x}{v_0\cos 45°}$ 

 

Using the kinematics equation in vertical direction,

 

 $$\begin{gathered} y=\left(v_0\sin 45°\right)\left(\frac{x}{v_0\cos 45°}\right)+\frac{1}{2}\left(-g\right){\left(\frac{x}{v_0\cos 45°}\right)}^2 \\ =x\tan 45°-\frac{1}{2}\frac{g{x}^2}{v_0^2\cos 45°} \end{gathered}$$

Now rearrange the equation for  $v_0$

 

$$\begin{gathered} \frac{1}{2}\frac{g{x}^2}{v_0^2{\cos }^{2}45°}=x \tan 45°-y \\ v_0^2=\frac{g{x}^2}{2\left(x \tan 45°-y\right){\cos }^{2}45°} \\ v_0=\sqrt{\frac{g{x}^2}{2\left(x \tan 45°-y\right){\cos }^{2}45°}} \end{gathered}$$ 

 

Substitute 10.9728m for x , 3.048m for y and 9.8$m/s^2$ for g in the above equation.

 

$$\begin{gathered} v_0=\sqrt{\frac{\left(9.8\right){\left(10.9728\right)}^2}{2\left(\left(10.9728\right)\tan 45°-3.048\right){\cos }^{2}45°}} \\ =12.2 m/s \end{gathered}$$

 

Convert the speed into

$$\begin{gathered} v_0=\left(12.2m/s\right)\left(\frac{3600s}{1h}\right)\left(\frac{1km}{1000m}\right) \\ =43.9km/h \end{gathered}$$ 

 

Therefore the speed of the ball is 43.9 km/h .