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Chapter 13: Problem 7

Firing golf balls 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
Initial speed is 9.9 m/s; angles for 6 m range are 18.65掳 and 71.35掳.

Step by step solution

01

Identify known values

We are given that the angle of launch of the golf ball is \( 45^{\circ} \) and the range is 10 meters. We need to find the initial speed. Also given is that for the range of 6 meters, we need to find the two angles with the same initial speed.
02

Set up range formula for projectile motion

The range \( R \) of a projectile launched with an initial speed \( v_0 \) at an angle \( \theta \) is given by the equation: \[ R = \frac{v_0^2 \sin(2\theta)}{g} \] where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.
03

Calculate initial speed for angle \( 45^{\circ} \)

Substitute \( R = 10 \), \( \theta = 45^{\circ} \), and \( g = 9.8 \) into the range formula: \[ 10 = \frac{v_0^2 \sin(90^{\circ})}{9.8} \] Since \( \sin(90^{\circ}) = 1 \), this becomes: \[ 10 = \frac{v_0^2}{9.8} \] Solve for \( v_0 \):\[ v_0^2 = 98 \] \[ v_0 = \sqrt{98} \approx 9.9 \, \text{m/s} \]
04

Set the range equation for a 6 m range

For the same initial speed \( v_0 \), set \( R = 6 \):\[ 6 = \frac{(9.9)^2 \sin(2\theta)}{9.8} \]
05

Solve for the angles

First, calculate \( \frac{{(9.9)^2}}{{9.8}} \approx 9.9 \), then:\[ 6 = 9.9 \sin(2\theta) \] \[ \sin(2\theta) = \frac{6}{9.9} \approx 0.606 \] Use the inverse sine function to find \( 2\theta \):\[ 2\theta = \arcsin(0.606) \approx 37.3^{\circ} \text{ or } 142.7^{\circ} \] \( \theta \) is then:\[ \theta = \frac{37.3}{2} \approx 18.65^{\circ} \text{ or } \frac{142.7}{2} \approx 71.35^{\circ} \]

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Key Concepts

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

Initial Speed Calculation
To find the initial speed of a projectile, like a golf ball fired from a spring gun, it's necessary to understand how speed affects the motion. Initial speed is the speed at which the projectile is launched into the air. It influences how far the projectile will travel. For accurate calculations, the initial speed \( v_0 \) can be deduced using the range formula of projectile motion. Given a problem where the golf ball lands 10 meters away at a launch angle of \( 45^{\circ} \), you can plug these known values into the formula. This specific angle simplifies calculations, as \( \sin(90^{\circ}) = 1 \). Substitute and solve the equation \[ 10 = \frac{v_0^2}{9.8} \] to find that \( v_0 \) is approximately \( 9.9 \, \text{m/s} \). Understanding this concept helps in predicting a projectile's motion, allowing control over how far it will go based on its initial speed.
Projectile Range Formula
The range formula for projectile motion is crucial in understanding how far a projectile will travel after it is launched. This formula takes into account several factors: the initial speed, the angle of launch, and gravity's effect.The formula is written as:\[ R = \frac{v_0^2 \sin(2\theta)}{g} \]Where:
  • \( R \) is the range, or how far the projectile lands away from its launch point.
  • \( v_0 \) is the initial speed.
  • \( \theta \) is the launch angle.
  • \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \).
Using this formula, you can calculate the possible range of a projectile given its speed and angle. It is with this formula that you initially calculate the speed necessary for a specific range and subsequently use it to determine potential launch angles for different desired ranges.
Angle of Launch
The angle of launch significantly influences a projectile's trajectory. Understanding how different angles affect range is key to mastering projectile motion problems.For a range of 10 meters with an angle of \( 45^{\circ} \), we already found an initial speed. But what if the range needs to be different? Consider the challenge of achieving a 6-meter range using the same speed. This requires solving for two angles.Substituting into the modified range equation:\[ 6 = \frac{(9.9)^2 \sin(2\theta)}{9.8} \]Rearranging and solving gives \( \sin(2\theta) = 0.606 \). Finding the \( \arcsin \) of this value results in two possible values for \( 2\theta \); either \( 37.3^{\circ} \) or \( 142.7^{\circ} \). Half these values to find practical launch angles: \( 18.65^{\circ} \) and \( 71.35^{\circ} \). This dual-solution occurs due to the trigonometric properties of sine, where both a lower and higher angle can yield the same sine value, thus affecting the range similarly.

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Most popular questions from this chapter

In Example \(5,\) we found the curvature of the helix \(\mathbf{r}(t)=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}+b t \mathbf{k}\) \((a, b \geq 0)\) to be \(\kappa=a /\left(a^{2}+b^{2}\right) .\) What is the largest value \(\kappa\) can have for a given value of \(b ?\) Give reasons for your answer.

As mentioned in the text, the tangent line to a smooth curve \(\mathbf{r}(t)=f(t) \mathbf{i}+g(t) \mathbf{j}+h(t) \mathbf{k}\) at \(t=t_{0}\) is the line that passes through the point \(\left(f\left(t_{0}\right), g\left(t_{0}\right), h\left(t_{0}\right)\right)\) parallel to \(\mathbf{v}\left(t_{0}\right),\) the curve's velocity vector at \(t_{0} .\) In Exercises \(33-36,\) find parametric equations for the line that is tangent to the given curve at the given parameter value \(t=t_{0}\) . $$ \mathbf{r}(t)=(a \sin t) \mathbf{i}+(a \cos t) \mathbf{j}+b t \mathbf{k}, \quad t_{0}=2 \pi $$

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