Chapter 10, Problem 1P

Solution

(a)

To find: Write an expression of the half of the total horizontal distance.

  $y$ coordinate is given.

Given information:

It is given that the horizontal distance $x$ travelled by a ball kicked from the ground level at speed of $v$ and angle $\theta$ can be modelled by $x=\frac{v^2}{32}\sin 2\theta$ .

Now half of the horizontal distance will be:

  $\begin{array}{l}x=\frac{1}{2}\cdot \frac{v^2}{32}\sin 2\theta \\ x=\frac{v^2}{64}\sin 2\theta \end{array}$

The ball travels in parabolic path, the maximum height occurs at half the horizontal distance.

(b)

To find: Write a simplified expression of the maximum height of the football.

The expression for the maximum height of the football is $y=\frac{v^2{\sin }^2\theta }{64}$ .

Given information:

It is given that half of the horizontal distance $x$ travelled by a ball kicked from the ground level at speed of $v$ and angle $\theta$ can be modelled by $x=\frac{v^2}{64}\sin 2\theta$ .

The projectile motion of the football is given by the equation:

  $y=\frac{-16}{v^2{\cos }^2\theta }{x}^2+\left(\tan \theta \right)x+h_0$

Substitute $x=\frac{v^2}{64}\sin 2\theta$ in the above equation with $h_0=0$ gives:

  $\begin{array}{l}y=\frac{-16}{v^2{\cos }^2\theta }{\left(\frac{v^2}{64}\sin 2\theta \right)}^2+\left(\tan \theta \right)\left(\frac{v^2}{64}\sin 2\theta \right)+0\\ y=-\frac{v^2\left(\sin 2\theta \right)\left(\sin 2\theta \right)}{256{\cos }^2\theta }+\left(\tan \theta \right)\left(\frac{v^2}{64}\sin 2\theta \right)\\ y=\frac{-v^2\left(2\sin \theta \cos \theta \right)\left(2\sin \theta \cos \theta \right)}{256{\cos }^2\theta }+\frac{\sin \theta }{\cos \theta }\left(\frac{v^2}{64}\left(2\sin \theta \cos \theta \right)\right)\\ y=\frac{-v^2{\sin }^2\theta }{64}+\frac{v^2{\sin }^2\theta }{32}\\ y=\frac{v^2{\sin }^2\theta }{64}\end{array}$

This is the required expression.

(c)

To find: Find the maximum height when $v=80,\theta =35$

The maximum height IS 33 feet.

Given information:

The expression for the maximum height of the football is $y=\frac{v^2{\sin }^2\theta }{64}$ .

Substitute the $v=80,\theta =35$ in $y=\frac{v^2{\sin }^2\theta }{64}$

  $\begin{array}{l}y=\frac{{80}^2{\sin }^2\left(35\right)}{64}\\ y=\frac{6400{\left(0.57\right)}^2}{64}\\ y=100\cdot 0.328\\ y\approx 33\end{array}$

This is the maximum height.

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