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Chapter 13: Problem 23

A spring gun at ground level fires a golf ball at an angle of \(45^{\circ} .\) The ball lands \(10 \mathrm{m}\) away. a. What was the ball's initial speed? b. For the same initial speed, find the two firing angles that make the range \(6 \mathrm{m}\)

Short Answer

Expert verified
a. The initial speed was \(9.9 \text{ m/s}\). b. The two angles are approximately \(18.44^{\circ}\) and \(71.56^{\circ}\).

Step by step solution

01

Understand the Problem

We have a projectile motion problem where a golf ball is fired at an angle of \(45^{\circ}\) and lands \(10 \text{ m}\) away. We need to find the initial speed for this scenario and the angles for the same initial speed that result in a \(6 \text{ m}\) range.
02

Use Projectile Motion Formula

The horizontal range \(R\) of a projectile is given by the formula \( R = \frac{v^2\sin(2\theta)}{g} \), where \(v\) is the initial speed, \(\theta\) is the launch angle, and \(g = 9.8 \text{ m/s}^2\) is the acceleration due to gravity. Since \(\theta = 45^{\circ}\), we have \( \sin(90^{\circ}) = 1 \).
03

Calculate Initial Speed

Plugging the values for range \(R = 10 \text{ m}\) and \(\sin(90^{\circ}) = 1\) into the formula gives us \(10 = \frac{v^2}{9.8}\). Solving for \(v^2\), we get \(v^2 = 10 \times 9.8\). Thus, \(v = \sqrt{98} = 9.9 \text{ m/s}\).
04

Determine Firing Angles for 6 m Range

For the range \(R = 6 \text{ m}\), we use the same formula with \( \sin(2\theta) = \frac{6 \times 9.8}{v^2} = \frac{58.8}{98} = 0.6\). Solving for \(2\theta\) gives \(2\theta = \sin^{-1}(0.6)\).
05

Solve for Two Angles

Calculating the inverse sine, \(2\theta = 36.87^{\circ}\) or \(2\theta = 180^{\circ} - 36.87^{\circ} = 143.13^{\circ}\). Therefore, \(\theta = 18.44^{\circ}\) or \(\theta = 71.56^{\circ}\).
06

Verify Calculations

Check calculated angles and speed back in the formula to ensure consistency with the intended range. Both angles when used provide the range of \(6 \text{ m}\), confirming calculations are correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial Speed Calculation
Initial speed is a crucial aspect of projectile motion problems. In our scenario, a golf ball is fired at a 45-degree angle and lands 10 meters away. To determine the initial speed, we use the formula for the range of a projectile:
  • \( R = \frac{v^2\sin(2\theta)}{g} \)
Here, \( R \) represents the range, \( \theta \) is the launch angle (45 degrees in this case), \( v \) is the initial speed, and \( g \) is the acceleration due to gravity (9.8 m/s虏). We know that \( \sin(90掳) = 1 \), simplifying the equation to:
  • \( 10 = \frac{v^2}{9.8} \)
By solving for \( v^2 \), we multiply 10 by 9.8, leading to \( v^2 = 98 \). Taking the square root provides the initial speed: \( v = \sqrt{98} = 9.9 \) m/s.
Launch Angle Determination
Launch angles are pivotal when examining projectile motion, as they decide the trajectory path. Here, with a determined initial speed of 9.9 m/s, calculating the appropriate angles to achieve different ranges is essential.
  • For a given speed, multiple angles can provide a projectile with the same horizontal distance. This principle arises from the symmetrical properties of sine function within its range constraints.
In our exercise, the challenge was to identify launch angles that yield a range of 6 meters. For this, we used the same range formula but solved for \( \theta \). This procedure involves closely engaging with the properties of trigonometric sine function, specifically using its periodicity and symmetry.
Range Formula
The range formula is a foundational equation in projectile motion. It's given by:
  • \( R = \frac{v^2\sin(2\theta)}{g} \)
This formula connects four critical variables:
  • The range \( R \),
  • Initial speed \( v \),
  • The angle \( \theta \),
  • And the gravitational constant \( g \), typically 9.8 m/s虏 on Earth.
Notably, sine of double the launch angle, \( \sin(2\theta) \), emphasizes trigonometric characteristics unique to projectile calculations. By imposing specific values to three of the variables, it's straightforward to solve for the fourth. Understanding and manipulating this formula allows students to tackle a wide array of projectile motion problems.
Inverse Sine Function
The inverse sine function, or arcsine, is an indispensable tool in resolving projectile motion issues. Specifically, it enables students to determine angles from known sine values, which is crucial for our exercise's second part. Here, we calculated the angles required for a range of 6 meters.
  • The inverse sine function, denoted \( \sin^{-1}(x) \), returns an angle whose sine value is \( x \).
In this case, after determining \( \sin(2\theta) = 0.6 \), using the inverse sine function gave us:
  • \( 2\theta = \sin^{-1}(0.6) = 36.87^{\circ} \),
  • or considering sine's symmetry, \( 2\theta = 180^{\circ} - 36.87^{\circ} = 143.13^{\circ} \).
Thus, the possible launch angles are \( \theta = 18.44^{\circ} \) and \( \theta = 71.56^{\circ} \), both of which allow the projectile to travel 6 meters.

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Most popular questions from this chapter

\(\mathbf{r}(t)\) is the position of a particle in space at time \(t\) Find the particle's velocity and acceleration vectors. Then find the particle's speed and direction of motion at the given value of \(t .\) Write the particle's velocity at that time as the product of its speed and direction. $$\mathbf{r}(t)=(2 \ln (t+1)) \mathbf{i}+t^{2} \mathbf{j}+\frac{t^{2}}{2} \mathbf{k}, \quad t=1$$

Give the position vectors of particles moving along various curves in the \(x y\) -plane. In each case, find the particle's velocity and acceleration vectors at the stated times and sketch them as vectorson the curve. $$\mathbf{r}(t)=\left(4 \cos \frac{t}{2}\right) \mathbf{i}+\left(4 \sin \frac{t}{2}\right) \mathbf{j} ; \quad t=\pi \text { and } 3 \pi / 2$$

Find the point on the curve $$\mathbf{r}(t)=(5 \sin t) \mathbf{i}+(5 \cos t) \mathbf{j}+12 t \mathbf{k}$$ at a distance \(26 \pi\) units along the curve from the point (0,5,0) in the direction of increasing arc length.

Find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. $$\mathbf{r}(t)=t \mathbf{i}+(2 / 3) t^{3 / 2} \mathbf{k}, \quad 0 \leq t \leq 8$$

Each of the following equations in parts (a)-(e) describes the motion of a particle having the same path, namely the unit circle \(x^{2}+y^{2}=1 .\) Although the path of each particle in parts (a)-(e) is the same, the behavior, or "dynamics," of each particle is different. For each particle, answer the following questions. i) Does the particle have constant speed? If so, what is its constant speed? ii) Is the particle's acceleration vector always orthogonal to its velocity vector? iii) Does the particle move clockwise or counterclockwise around the circle? iv) Does the particle begin at the point (1,0)\(?\) a. \(\mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}, \quad t \geq 0\) b. \(\mathbf{r}(t)=\cos (2 t) \mathbf{i}+\sin (2 t) \mathbf{j}, \quad t \geq 0\) c. \(\mathbf{r}(t)=\cos (t-\pi / 2) \mathbf{i}+\sin (t-\pi / 2) \mathbf{j}, \quad t \geq 0\) d. \(\mathbf{r}(t)=(\cos t) \mathbf{i}-(\sin t) \mathbf{j}, \quad t \geq 0\) e. \(\mathbf{r}(t)=\cos \left(t^{2}\right) \mathbf{i}+\sin \left(t^{2}\right) \mathbf{j}, \quad t \geq 0\)
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