Chapter 5, Problem 6PS

To determine

Find a formula for the maximum height of an object projected from ground level at velocity $v_0$ and angle $\theta$ .

Answer to Problem 6PS

  $\boxed{\frac{v_o{}^2{\sin }^2\theta }{64}}$

Explanation of Solution

Given information:

The path travelled by an object (neglecting air resistance) that is projected at an initial height of $h_0$ feet, an initial velocity of $v_0$ feet per second, and an initial angle $\theta$ is given by

  $y=-\frac{16}{v_0{}^2{\cos }^2\theta }{x}^2+\left(\tan \theta \right)x+h_0$

where the horizontal distance $x$ and the vertical distance $y$ are measured in feet. Find a formula for the maximum height of an object projected from ground level at velocity $v_0$ and angle $\theta$ . To do this, find half of the horizontal distance

  $\frac{1}{32}{v}_0{}^2\sin 2\theta$

And then substitute it for $x$ in the model for the path of a projectile (where $h_0=0$ ).

Calculation:

Consider the following model for the path travelled by an object,

  $y=-\frac{16}{v_0{}^2{\cos }^2\theta }{x}^2+\left(\tan \theta \right)x+h_0$

The object is projected at an initial height $h$ , velocity $v_0$ and angle $\theta$

Here $x$ and $y$ are measured in feet.

The half of the horizontal distance of $\frac{1}{32}{v}_0{}^2\sin 2\theta$ is,

  $\frac{1}{2}\times \left(\frac{1}{32}{v}_0{}^2\sin 2\theta \right)=\frac{1}{64}{v}_0{}^2\sin 2\theta$

Now, put the value of horizontal half distance and $h_0=0$ in the general model,

  $\begin{array}{l}y=-\frac{16}{v_0{}^2{\cos }^2\theta }{\left(\frac{1}{64}{v}_o{}^2\sin 2\theta \right)}^2+\left(\tan \theta \right)\left(\frac{1}{64}{v}_o{}^2\sin 2\theta \right)+h_0\\ =-\frac{\cancel{16}}{\cancel{v_0{}^2}\cancel{{\cos }^2\theta }}\times \frac{1}{\cancel{4096}}\cancel{v_0{}^2}4{\sin }^2\theta \cancel{{\cos }^2\theta }+\left(\frac{\sin \theta }{\cancel{\cos \theta }}\right)\left(\frac{1}{\cancel{64}}{v}_0{}^2\cancel{2}\sin \theta \cancel{\cos \theta }\right)+0\\ =-\frac{1}{\cancel{256}}\times \cancel{4}{v}_0{}^2{\sin }^2\theta +\frac{1}{32}{v}_o{}^2{\sin }^2\theta \\ =-\frac{1}{64}{v}_0{}^2{\sin }^2\theta +\frac{1}{32}{v}_0{}^2{\sin }^2\theta \end{array}$

Simplify at follows,

  $\begin{array}{l}y=-\frac{1}{64}{v}_0{}^2{\sin }^2\theta +\frac{1}{32}{v}_0{}^2{\sin }^2\theta \\ =\frac{-v_0{}^2{\sin }^2\theta +2v_0{}^2{\sin }^2\theta }{64}\\ =\frac{v_0{}^2{\sin }^2\theta }{64}\end{array}$

Hence, the formula for maximum height of an object projected from the ground level is,

  $\boxed{\frac{v_o{}^2{\sin }^2\theta }{64}}$