Problem 58 - Calculus Early Transcendentals: Pearson New International Edition

Question

The path of a projectile fired from level ground with a speed of \(v_{0}\) feet per second at an angle \(\alpha\) with the ground is given by the parametric equations $$ x=\left(v_{0} \cos \alpha\right) t, \quad y=-16 t^{2}+\left(v_{0} \sin \alpha\right) t $$ (a) Show that the path is a parabola. (b) Find the time of flight. (c) Show that the range (horizontal distance traveled) is \(\left(v_{0}^{2} / 32\right) \sin 2 \alpha\). (d) For a given \(v_{0}\), what value of \(\alpha\) gives the largest possible range?

Step-by-Step Solution

Answer

The path is a parabola; time of flight is \(\frac{v_0 \sin \alpha}{16}\); range is \(\frac{v_0^2}{32} \sin 2\alpha\); max range at \(\alpha = 45^\circ\).

Step 1 - Understanding the parametric equations

The parametric equations of the projectile motion are given as: \[ x = (v_{0} \cos \alpha) t \] \[ y = -16t^{2} + (v_{0} \sin \alpha) t \] These equations describe a projectile's motion under gravity, where the horizontal position is determined by time and the horizontal velocity component, whereas the vertical position depends on time, initial vertical velocity, and gravitational pull.

Step 2 - Show the path is a parabola

Eliminate the parameter \( t \) from these equations to find the relationship between \( x \) and \( y \). From \( x = (v_{0} \cos \alpha)t \), solve for \( t \): \[ t = \frac{x}{v_{0} \cos \alpha} \] Substitute \( t \) into the equation for \( y \):\[ y = -16\left( \frac{x}{v_{0} \cos \alpha} \right)^2 + (v_{0} \sin \alpha) \left( \frac{x}{v_{0} \cos \alpha} \right) \] Simplifying this, we obtain:\[ y = -\frac{16x^2}{(v_0^2 \cos^2 \alpha)} + x \tan \alpha \]This is a quadratic equation in terms of \( x \), which confirms that the path is a parabola.

Step 3 - Find the time of flight

The flight ends when the projectile returns to the ground, i.e., when \( y = 0 \).Set the equation for \( y = -16t^2 + (v_0 \sin \alpha)t \) equal to zero:\[ -16t^2 + (v_0 \sin \alpha)t = 0 \]Factor out \( t \):\[ t(-16t + v_0 \sin \alpha) = 0 \]This gives solutions \( t = 0 \) (launch time) and \( t = \frac{v_0 \sin \alpha}{16} \) (time of flight).The time of flight is \( \frac{v_0 \sin \alpha}{16} \).

Step 4 - Show the range formula

The range is the horizontal distance traveled after the time of flight. Plug \( t = \frac{v_0 \sin \alpha}{16} \) into \( x = (v_0 \cos \alpha)t \):\[ x = (v_0 \cos \alpha) \left( \frac{v_0 \sin \alpha}{16} \right) \]Simplify:\[ x = \frac{v_0^2 \cos \alpha \sin \alpha}{16} \]Using the trigonometric identity \( 2 \sin \alpha \cos \alpha = \sin 2\alpha \), the range is:\[ x = \frac{v_0^2}{32} \sin 2\alpha \]This confirms the range formula.

Step 5 - Find the angle for maximum range

To maximize the range \( x = \frac{v_0^2}{32} \sin 2\alpha \), the value of \( \sin 2\alpha \) should be maximum, which is 1 (since max value of sine is 1).\( \sin 2\alpha = 1 \) when \( 2\alpha = 90^\circ \).Thus, \( \alpha = 45^\circ \).The largest possible range is achieved when the launch angle is \( \alpha = 45^\circ \).

Key Concepts

Parametric Equations

In projectile motion, parametric equations are crucial for describing the path of an object in flight. They express the position coordinates \((x, y)\) in terms of a third variable, time \(t\). For a projectile fired from level ground, the equations are given by:

• Horizontal position: \(x = (v_{0} \cos \alpha)t\)
• Vertical position: \(y = -16t^2 + (v_{0} \sin \alpha)t\)

These equations let us separately handle horizontal and vertical motions. The first equation focuses on horizontal distance, driven by the component \(v_{0} \cos \alpha\). The second equation captures the vertical component, which is influenced by gravity \(-16t^2\) and the initial vertical velocity \((v_{0} \sin \alpha)t\). Understanding these components helps predict where the projectile will land.

Parabolic Motion

The trajectory of a projectile under the effect of gravity is parabolic. This parabola stems from the parametric equations where the vertical motion is subject to constant acceleration due to gravity, while horizontal motion has a constant velocity.
To prove the parabola shape, eliminate \(t\) from the equations. Substitute \(t = \frac{x}{v_{0} \cos \alpha}\) into the equation for \(y\), creating a relation between \(x\) and \(y\):\[ y = -\frac{16x^2}{(v_0^2 \cos^2 \alpha)} + x \tan \alpha \]This quadratic relationship confirms the path is parabolic. The negative coefficient of \(x^2\) demonstrates the typical downward curvature of a parabola due to gravity's effect on the projectile.

Trigonometric Identities

Trigonometric identities play a crucial role in simplifying expressions in projectile motion. Particularly, the identity \(2 \sin \alpha \cos \alpha = \sin 2\alpha\) transforms expressions, simplifying the calculations for range.
In the step for finding range, the formula simplifies using this identity:\[ x = \frac{v_0^2}{32} \sin 2\alpha \]By recognizing \( \sin 2\alpha \) as \( 2 \sin \alpha \cos \alpha \), we seamlessly transition from a longer expression to a compact form. Such identities often unlock simpler solutions, providing better insights into the problem, especially when analyzing angles and maximizing other parameters.

Maximize Range

Maximizing the range of a projectile is a common goal, and for a given initial velocity, this involves finding the optimal launch angle. The formula for range is given by:\[ x = \frac{v_0^2}{32} \sin 2\alpha \]To achieve maximum range, \(\sin 2\alpha\) should hit its peak value, which is 1. This occurs when \(2\alpha = 90^\circ\), thus making the ideal launch angle \(\alpha = 45^\circ\).
Launching at \(45^\circ\) essentially balances the vertical and horizontal components, providing the longest possible distance. Therefore, understanding this relationship helps optimize strategies where range is crucial, such as in sports or engineering applications where projectiles are used.