The ideas we are practicing in these exercises can be put to practical use. A calculus teacher's son likes to shoot arrows. He was wondering at what speed the arrow leaves the bow when it is released. So they went to a large deserted beach at the Great Salt Lake and shot the bow at an angle of 45 degrees. The arrow hit the ground at a distance of 600 feet. All of this was measured very roughly, but let's work with these figures, and let's ignore air resistance. For simplicity let's also assume that the arrow is released at an initial height of 0 feet. As usual, assume that gravity causes an object to accelerate at 32 feet per second squared.
The arrow leaves the bow at what speed in miles per hour?
To find the speed at which the arrow leaves the bow in miles per hour, we need to analyze the projectile motion of the arrow. Here's how we can do it:
Step 1: Convert the given distance of 600 feet into meters.
- 1 foot is approximately equal to 0.3048 meters.
- Therefore, 600 feet is equal to 600 * 0.3048 = 182.88 meters.
Step 2: Convert the given acceleration due to gravity from feet per second squared to meters per second squared.
- 1 foot is approximately equal to 0.3048 meters.
- Therefore, 32 feet per second squared is equal to 32 * 0.3048 = 9.81 meters per second squared.
Step 3: Determine the time of flight of the arrow.
- The time it takes for the arrow to reach the ground can be found using the horizontal distance and initial velocity.
- Since the arrow is shot at an angle of 45 degrees, the horizontal and vertical components of the initial velocity are equal.
- The horizontal distance traveled can be considered the range, which is 182.88 meters in this case.
- The formula for the range is: range = (initial velocity^2 * sin(2*angle))/gravity
- Plugging in the given values, the equation becomes: 182.88 = (initial velocity^2 * sin(90))/9.81
- Since sin(90) is equal to 1, the equation simplifies to: initial velocity^2 = 182.88 * 9.81
- Solving for the initial velocity, we get: initial velocity = sqrt(182.88 * 9.81).
Step 4: Convert the initial velocity from meters per second to miles per hour.
- 1 meter per second is equal to approximately 2.23694 miles per hour.
- Therefore, the initial velocity in miles per hour is calculated as follows: initial velocity (miles per hour) = sqrt(182.88 * 9.81) * 2.23694.
By following these steps, you will be able to calculate the speed at which the arrow leaves the bow in miles per hour.
To find the speed at which the arrow leaves the bow, we can use the equations of projectile motion. We know that the arrow was shot at an angle of 45 degrees and hit the ground at a distance of 600 feet. Let's denote the initial speed of the arrow as V.
Step 1: Break down the initial velocity into its horizontal and vertical components.
The initial velocity can be divided into its horizontal and vertical components using trigonometry. Since the angle of projection is 45 degrees, both components will have the same magnitude. Therefore, the horizontal and vertical components are:
Vx = V * cos(45)
Vy = V * sin(45)
Step 2: Determine the time it takes for the arrow to hit the ground.
Since we are assuming no air resistance, the time taken for the arrow to hit the ground will be the same for both the horizontal and vertical components. We can use the equation:
d = Vt + (0.5) * a * t^2
where d is the vertical distance (600 feet), t is the time, and a is gravity (-32 ft/s^2). Since the initial height is 0 feet, it can be ignored in this equation.
By substituting the values, the equation becomes:
600 = (0.5) * (-32) * t^2
Simplifying the equation:
300 = -16t^2
Divide both sides by -16:
t^2 = -300 / -16
t^2 = 18.75
Taking the square root of both sides:
t ≈ 4.33 seconds
Step 3: Determine the vertical component of the initial velocity.
Now that we have the time, we can use it to calculate the vertical component of the initial velocity. We can use the equation:
Vy = V * sin(θ) - gt
where θ is the angle of projection (45 degrees), g is gravity (-32 ft/s^2), and t is the time.
By substituting the values, the equation becomes:
V * sin(45) = V * sin(45) - (-32) * 4.33
Simplifying the equation:
0 = -4.33V + 139.52
Rearranging the equation:
4.33V = 139.52
Divide both sides by 4.33:
V ≈ 32.24 ft/s
Step 4: Convert the speed to miles per hour.
To convert the speed from feet per second to miles per hour, we can use the conversion factor:
1 mile = 5280 feet
1 hour = 3600 seconds
Therefore,
V_mph = (V_ft/s * 3600) / 5280
By substituting the value of V:
V_mph ≈ (32.24 * 3600) / 5280
Simplifying the equation:
V_mph ≈ 22.03 mph
So, the speed at which the arrow leaves the bow is approximately 22.03 miles per hour.