Chapter 10.7, Problem 60AYU

To determine

To find:

a. The rectangular equation of the trajectory and identify the curve.

Answer to Problem 60AYU

a. $y = x\text{tan}\theta - \frac{16x^2}{v_o^2\cos ²\theta }$

Explanation of Solution

Given:

The position of a projectile fired with an initial velocity $v_0$ feet per second and at an angle$\theta$ to the horizontal at the end of $t$ seconds is given by the parametric equations.

$x = (v_0\text{cos}\theta )t$, $y = \left(v_0\text{sin}\theta \right)t - 16t²$

Formula used:

$x = \left(v_o\text{cos}\theta \right)t$, $y = -\frac{1}{2}gt² + \left(v_o\text{sin}\theta \right)t + h$

Calculation:

a. $x = (v_0\text{cos}\theta )t$, $y = \left(v_0\text{sin}\theta \right)t - 16t²$

$t = x/v_o\text{cos}\theta$

$y = \left(v_o\text{sin}\theta \right)\left(\frac{x}{v_o\text{cos}\theta }\right) - 16\left(\frac{x}{v_o\text{cos}\theta }\right)²$

$y = x\text{tan}\theta - 16\left(\frac{x}{v_o\text{cos}\theta }\right)²$

$y = x\text{tan}\theta - \frac{16x^2}{v_o^2\cos ²\theta }$

$y$ is a quadratic function of $x$ and it represents the graph of a parabola where $a = -\frac{16}{v_o^2\cos \theta }$, $b = \text{tan}\theta$ and $c = 0$.

To determine

To find:

b. To show that the projectile hits the ground $\left(y = 0\right)$ when $t=1/16v_o\text{sin}\theta$.

Answer to Problem 60AYU

b. $t = \frac{v_o\text{sin}\theta }{16}$

Explanation of Solution

Given:

The position of a projectile fired with an initial velocity $v_0$ feet per second and at an angle$\theta$ to the horizontal at the end of $t$ seconds is given by the parametric equations.

$x = (v_0\text{cos}\theta )t$, $y = \left(v_0\text{sin}\theta \right)t - 16t²$

Formula used:

$x = \left(v_o\text{cos}\theta \right)t$, $y = -\frac{1}{2}gt² + \left(v_o\text{sin}\theta \right)t + h$

Calculation:

b. $y = 0$

$y = \left(v_o\text{sin}\theta \right)t - 16t² = 0$

$t\left(\left(v_o\text{sin}\theta \right) - 16t\right) = 0$

$t = \frac{v_o\text{sin}\theta }{16}$

To determine

To find:

c. How far has the projectile traveled (horizontally) when it strikes the ground? In other words, find the range R.

Answer to Problem 60AYU

c. $x = v_o²\text{sin}2\theta /32$

Explanation of Solution

Given:

The position of a projectile fired with an initial velocity $v_0$ feet per second and at an angle$\theta$ to the horizontal at the end of $t$ seconds is given by the parametric equations.

$x = (v_0\text{cos}\theta )t$, $y = \left(v_0\text{sin}\theta \right)t - 16t²$

Formula used:

$x = \left(v_o\text{cos}\theta \right)t$, $y = -\frac{1}{2}gt² + \left(v_o\text{sin}\theta \right)t + h$

Calculation:

c. To find R

We substitute $t$ in $x$

$x = \left(v_o\text{cos}\theta \right)t$

$x = \left(v_o\text{cos}\theta \right)\frac{v_o\text{sin}\theta }{16}$

$x = v_o²\text{cos}\theta \text{sin}\theta /16$

$x = v_o²\text{sin}2\theta /32$

To determine

To find:

d. Find the time $t$ when $x=y$. Then find the horizontal distance $x$ and the vertical distance $y$ traveled by the projectile in this time. Then compute $\surd x² + y²$. This is the distance R, the range, that the projectile travels up a plane inclined at $45°$ to the horizontal $\left(x = y\right)$.

Answer to Problem 60AYU

d. $\sqrt{x² + y^2} = \surd 2\left(\left(\frac{v_o^2\cos \theta }{16}\right)\left(\text{sin}\theta - \text{cos}\theta \right)\right)$ $\theta > 45$ as $\text{sin}\theta - \text{cos}\theta > 0$

Explanation of Solution

Given:

The position of a projectile fired with an initial velocity $v_0$ feet per second and at an angle$\theta$ to the horizontal at the end of $t$ seconds is given by the parametric equations.

$x = (v_0\text{cos}\theta )t$, $y = \left(v_0\text{sin}\theta \right)t - 16t²$

Formula used:

$x = \left(v_o\text{cos}\theta \right)t$, $y = -\frac{1}{2}gt² + \left(v_o\text{sin}\theta \right)t + h$

Calculation:

d. $x = y$

$\left(v_o\text{cos}\theta \right)t = \left(v_o\text{sin}\theta \right)t - 16t²$

$16t² + \left(v_o\text{cos}\theta \right)t - \left(v_o\text{sin}\theta \right)t = 0$

$t\left(16t + \left(v_o\text{cos}\theta \right) - \left(v_o\text{sin}\theta \right)\right) = 0$

$t = 0$; $t = \frac{v_o\text{sin}\theta - v_o\text{cos}\theta }{16}$

At $t = \frac{v_o\text{sin}\theta - v_o\text{cos}\theta }{16}$

$x = \left(v_o\text{cos}\theta \right)\frac{v_o\text{sin}\theta - v_o\text{cos}\theta }{16} = \left(\frac{v_o^2\text{cos}\theta }{16}\right)\left(\text{sin}\theta - \text{cos}\theta \right)$

$x = y$ hence,

$y = \left(\frac{v_o^2 - \cos ²\theta + \text{sin}\theta \text{cos}\theta }{16}\right)$

$\sqrt{x² + y^2} = \surd 2\left(\left(\frac{v_o^2\cos \theta }{16}\right)\left(\text{sin}\theta - \text{cos}\theta \right)\right)$ $\theta > 45$ as $\text{sin}\theta - \text{cos}\theta > 0$