Problem 75 - Precalculus with Calculus Previews

Question

We saw in Problem 56 of Exercises \(4.2\) that if a projectile, such as a shot put, is released from a height \(h\), upward at an angle \(\theta\) with velocity \(v_{\mathrm{o}}\), the range \(R\) at which it strikes the ground is given by $$ R=\frac{v_{0} \cos \theta}{g}\left(v_{0} \sin \theta+\sqrt{v_{0}^{2} \sin ^{2} \theta+2 g h}\right), $$ where \(g\) is the acceleration due to gravity. (a) Show that when \(h=0\) the range of the projectile is $$ R=\frac{v_{0}^{2} \sin 2 \theta}{g} $$ (b) It can be shown that the maximum range \(R_{\max }\) is achieved when the angle \(\theta\) satisfies the equation $$ \cos 2 \theta=\frac{g h}{v_{0}^{2}+g h} $$ Show that maximum range is $$ R_{\max }=\frac{v_{0} \sqrt{v_{0}^{2}+2 g h}}{g} $$ by using the expressions for \(R\) and \(\cos 2 \theta\) and the half-angle formulas for the sine and the cosine with \(t=2 \theta\).

Step-by-Step Solution

Answer

(a) Substitute \(h = 0\) to simplify. (b) Use \(\cos 2\theta\) to find \(R_{\max}\).

Step 1: Set Initial Height to Zero

For part (a), we need to show that the formula simplifies when the height \( h = 0 \). Substitute \( h = 0 \) into the original range formula:\[R = \frac{v_0 \cos \theta}{g} \left(v_0 \sin \theta + \sqrt{v_0^2 \sin^2 \theta + 2gh }\right) = \frac{v_0 \cos \theta}{g} \left(v_0 \sin \theta + \sqrt{v_0^2 \sin^2 \theta}\right)\]Since \( h = 0 \), \( 2gh \) becomes zero, leading to simplification.

Step 2: Simplify the Expression

Recognizing that \( \sqrt{v_0^2 \sin^2 \theta}\) simplifies to \( v_0 \sin \theta \), we have:\[R = \frac{v_0 \cos \theta}{g} \left(v_0 \sin \theta + v_0 \sin \theta \right)\]This becomes:\[R = \frac{v_0 \cos \theta}{g} \left(2 v_0 \sin \theta\right) = \frac{2v_0^2 \cos \theta \sin \theta}{g}\]

Step 3: Apply Double Angle Formula

Using the double angle identity \( \sin 2\theta = 2 \sin \theta \cos \theta \), substitute into the expression:\[R = \frac{v_0^2 \sin 2\theta}{g}\]This is the equation we need to show for part (a).

Step 4: Introduce Maximum Range Condition

For part (b), substituting the condition \( \cos 2\theta = \frac{gh}{v_0^2 + gh} \) into the maximum range equation involves starting from the range formula.We use the given relation and identity \( \cos 2\theta = 1 - 2 \sin^2 \theta \).

Step 5: Solve for \( \sin^2 \theta \)

Rearranging \( \cos 2\theta = \frac{gh}{v_0^2 + gh} = 1 - 2 \sin^2 \theta \), solve for \( \sin^2 \theta \):\[2 \sin^2 \theta = 1 - \frac{gh}{v_0^2 + gh }\]From this, we have:\[\sin^2 \theta = \frac{v_0^2 + gh - gh}{2(v_0^2 + gh)} = \frac{v_0^2}{2(v_0^2 + gh)}\]

Step 6: Obtain Maximum Range

Substitute \( \sin \theta \) back into the range expression:\[R_{\max} = \frac{v_0 \cos \theta}{g}\left(v_0 \sin \theta + \sqrt{v_0^2\sin^2\theta + 2gh}\right)\]With the half-angles, we know \( \sin \theta = \sqrt{\frac{v_0^2}{2(v_0^2 + gh)}} \) and simplify the term:\[\cos \theta = \sqrt{1 - \sin^2 \theta}\]Substitute \( \cos \theta \) and \( \sin \theta \) to verify:\[R_{\max} = \frac{v_0 \sqrt{v_0^2 + 2gh}}{g}\] This uses well-established trigonometric identities.

Key Concepts

Range Formula

The range formula for projectile motion is a foundational concept in physics that allows us to calculate how far a projectile will travel horizontally. The general equation for the range \( R \) of a projectile launched from a height \( h \) with an initial velocity \( v_0 \) at an angle \( \theta \) is:
\[R = \frac{v_0 \cos \theta}{g} \left(v_0 \sin \theta + \sqrt{v_0^2 \sin^2 \theta + 2gh} \right)\]Here, \( g \) stands for the acceleration due to gravity.This formula is derived from integrating the components of motion:

• \( v_0 \cos \theta \) represents the horizontal component of motion.
• Extra terms under the square root account for the initial launch height, \( h \).
• The entire term is divided by \( g \) to account for gravitational effects.

Understanding this formula is crucial as it combines aspects of kinematics and dynamics to evaluate projectile motion comprehensively.

Maximum Range

When launching a projectile, determining the maximum range is about finding the optimal angle that yields the farthest distance. For a projectile launched from the ground (\( h = 0 \)), the angle \( \theta \) that will give you the maximum range is \( 45^\circ \). However, when launched from a height, the angle is slightly less.Using the condition for maximization:
\[\cos 2\theta = \frac{gh}{v_0^2 + gh}\]We derive the expression for maximum range as:
\[R_{\max} = \frac{v_0 \sqrt{v_0^2 + 2gh}}{g}\]This formula reveals:

• Increased velocity \( v_0 \) or greater initial height \( h \) will increase \( R_{\max} \).
• The calculation respects the conservation of energy, balancing potential and kinetic energy changes.

Achieving maximum range is about utilizing these physics principles to optimize launch conditions.

Trigonometric Identities

Trigonometric identities play a key role in simplifying complex projectile motion problems. Specifically, identities help relate angles to the properties of right triangles, which directly inform projectile calculations.Two crucial identities in this topic are:

• Double Angle Formula: \( \sin 2\theta = 2 \sin \theta \cos \theta \)
• Expression for cosine: \( \cos 2\theta = 1 - 2 \sin^2 \theta \)

Using these identities, we can reduce and manipulate the range formula to reach more concise forms. For example, these identities convert the complex trigonometric components into algebraically solvable forms, revealing deeper insight into projectile dynamics and simplifying calculations at various steps.

Acceleration Due to Gravity

The constant \( g \) for acceleration due to gravity is a pivotal factor in projectile motion. On Earth, \( g \) typically stands at \( 9.8 \, \text{m/s}^2 \), providing a baseline to account for gravitational pull.In projectile motion:

• \( g \) influences both vertical motion and ultimately the horizontal range.
• By understanding effects of \( g \), one predicts time of flight, max height, and landing speed.

Additionally, changes in \( g \) (e.g., on other planets) would significantly alter projectile behavior, prompting recalculations for each scenario. Recognizing the role of gravity thus ensures accuracy in projectile path predictions, whether in terrestrial contexts or interplanetary scenarios.