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Chapter 3: Problem 99

You want to make a plot of the trajectory of a projectile. That is, you want to make a plot of the height \(y\) of the projectile as a function of horizontal distance \(x\). The projectile is launched from the origin with an initial speed \(v_{\mathrm{i}}\) at an angle \(\theta\) above the horizontal. Show that the equation of the trajectory followed by the projectile is $$y=\left(\frac{v_{\text {iy }}}{v_{\text {ix }}}\right) x+\left(\frac{-g}{2 v_{\text {ix }}^{2}}\right) x^{2}$$

Short Answer

Expert verified
Question: Derive the trajectory equation for a projectile launched at an angle θ above the horizontal with an initial speed of 𝑣𝑖. Answer: The trajectory equation for a projectile launched at an angle θ above the horizontal with an initial speed of 𝑣𝑖 is given by \(y=\left(\frac{v_{\text{iy}}}{v_{\text{ix}}}\right) x+\left(\frac{-g}{2v_{\text{ix }}^{2}}\right) x^{2}\), where \(v_{\text{ix}} = v_\mathrm{i} \cos \theta\), and \(v_{\text{iy}} = v_\mathrm{i} \sin \theta\).

Step by step solution

01

Find the horizontal and vertical components of the initial velocity

To find the horizontal and vertical components of the initial velocity, we use the trigonometric functions, sine and cosine, multiplied with the initial speed. These components are given by: $$ v_{\text{ix}} = v_\mathrm{i} \cos \theta \\ v_{\text{iy}} = v_\mathrm{i} \sin \theta $$
02

Write the kinematic equations for horizontal and vertical motion

For the horizontal motion, there is no acceleration, and the equation is given by: $$ x = v_{\text{ix}} t $$ For the vertical motion, the acceleration is the gravitational acceleration \(g\), and we have the following equation: $$ y = v_{\text{iy}}t - \frac{1}{2}gt^2 $$
03

Substitute the time from the horizontal motion equation into the vertical motion equation

Rearrange the horizontal motion equation to isolate \(t\): $$ t=\frac{x}{v_{\text{ix}}} $$ Now, substitute the expression for \(t\) into the vertical motion equation: $$ y = v_{\text{iy}} \frac{x}{v_{\text{ix}}} - \frac{1}{2}g \left(\frac{x}{v_{\text{ix}}}\right)^2 $$
04

Simplify and rewrite the trajectory equation

Simplify the expression to get the desired trajectory equation: $$ y=\left(\frac{v_{\text {iy }}}{v_{\text {ix }}}\right) x+\left(\frac{-g}{2 v_{\text {ix }}^{2}}\right) x^{2} $$ This is the trajectory equation for a projectile launched at an angle \(\theta\) above the horizontal with initial speed \(v_{\mathrm{i}}\).

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