Question

Ask you to solve a variety of problems based on the principles of projectile motion. A hunter aims at a deer which is 40 yards away. Her crossbow is at a height of \(5 \mathrm{ft}\), and she aims for a spot on the deer \(4 \mathrm{ft}\) above the ground. The crossbow fires her arrows at \(300 \mathrm{ft} / \mathrm{s}\) (a) At what angle of elevation should she hold the crossbow to hit her target? (b) If the deer is moving perpendicularly to her line of sight at a rate of \(20 \mathrm{mph}\), by approximately how much should she lead the deer in order to hit it in the desired location?

Short answer

(a) Approximately \(0.383^\circ\) angle of elevation. (b) Lead the deer by about 23.55 feet.

Step-by-step solution

Calculate the time of flight

First, we need to find how long it takes for the arrow to travel across the horizontal distance to the target which is 40 yards away. We convert 40 yards to feet: \(40 ext{ yards} = 120 ext{ feet}.\) We then use the equation of motion horizontally which relates initial velocity, angle of projection \(\theta\) and time:\[x = v_0 \cos(\theta) \, t\]Substituting the given values: \(120 = 300 \cos(\theta) \, t\). Solving for \(t\), we have:\[ t = \frac{120}{300 \cos(\theta)} = \frac{2}{5 \cos(\theta)} \text{ seconds}.\]

Use the vertical motion to find the angle of projection

Using the vertical motion equation:\[y = v_0 \sin(\theta) \, t - \frac{1}{2}gt^2\]We set \(y = 4 - 5 = -1 \text{ foot} \) (since the arrow starts from 5 feet and should hit 4 feet, it slightly drops). Substituting the values,\[-1 = 300 \sin(\theta) \left(\frac{2}{5 \cos(\theta)}\right) - \frac{1}{2}(32.2)\left(\frac{2}{5 \cos(\theta)}\right)^2\].Solving the above equation gives us the angle \(\theta\). After simplification, the equation becomes:\[\tan(\theta) = 0.0067\]Solving this will give approximately\[\theta \approx 0.383^\circ.\]

Find the distance to lead the deer

Now, for part (b), we will calculate how much lead is needed for a moving deer. The deer moves at a speed of \(20 \text{ mph}\). Converting this to feet per second, we have:\[20 \text{ mph} = \frac{20 \times 5280}{3600} \approx 29.33 \text{ ft/s}.\]The time of flight was calculated to be approximately \(\frac{2}{5 \cos(0.383^\circ)} \approx 0.803 \text{ s}.\)The lead distance can be found using:\[\text{Lead Distance} = \text{Deer's Speed} \times \text{Time of Flight} = 29.33 \times 0.803 \approx 23.55 \text{ feet}.\]So, approximately 23.55 feet is how much the hunter should lead the deer.

Key concepts

Velocity Calculations

Understanding how to calculate velocity in projectile motion is crucial. It involves breaking down the velocity into horizontal and vertical components. When you fire an object, like an arrow, the initial velocity (\(v_0\)) must be split into two directions:

• Horizontal component (\(v_{0x} = v_0 \cos(\theta)\))
• Vertical component (\(v_{0y} = v_0 \sin(\theta)\))

These components help in determining the arrow's trajectory and impact point.
The speed at which an arrow travels affects how far and how fast it reaches the target. Therefore, when calculating these components, you must know the initial velocity and the angle at which the projectile is launched.
In our exercise, we see the arrow's speed is 300 ft/s. Being able to break down this speed using trigonometric functions is a key skill in solving projectile motion problems.

Angle of Elevation

The angle of elevation is the angle between the horizontal line and the line of sight to the object being targeted.
This angle is vital because it decides how the projectile will ascend and descend. In our problem, the solution involves finding this angle to ensure the arrow hits the target 4 feet above ground level.

To determine this angle, we used vertical motion equations:

• Equation: \(y = v_0 \sin(\theta) \, t - \frac{1}{2}gt^2\)
• The height difference between start and target is \(-1\) foot

Solving these equations helps find \(\theta\), where\(\tan(\theta) = 0.0067\), leading to an angle of roughly \(0.383^\circ\).
This tiny degree illustrates the precision needed in real-world applications like hunting or sports.

Time of Flight

Time of Flight refers to the duration a projectile is in the air from launch to impact.
In projectile motion problems, determining this time is crucial as it affects other calculations like distance or target lead.
Using our example, we calculated the time for the arrow to reach a 120-foot target:

• Instinctively, the formula \(t = \frac{x}{v_0 \cos(\theta)}\) calculates flight time.

This equation relies on knowing how far the target is, the initial speed, and the correct elevation angle.
The result \(t \approx 0.803\) seconds means that the arrow travels for almost a second before hitting.
This short time emphasizes the need for quick and precise adjustments during projectile launches.

Trigonometry in Physics

Trigonometry plays a significant role in physics, particularly in analyzing projectile motion. It aids in resolving complex motions into simpler components that can be studied easily.
In our exercise, the trigonometric functions \(\sin\) and \(\cos\) helped break down velocity into horizontal and vertical parts.

This breakdown allows separate analysis of

• Horizontal motion (with constant velocity)
• Vertical motion (affected by gravity)

Using identities like \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\) also allowed us to find the angle needed for precise aiming.
Thus, by using trigonometry effectively, we can solve many real-world problems relating to height, distance, speed, and time in physics.