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 a speed of how many miles per hour?
To find the speed of the arrow when it leaves the bow, we can use the kinematic equation for projectile motion. In this case, we can use the equation that relates the horizontal distance traveled (range) and the initial velocity of the arrow.
First, let's convert the given distance of 600 feet to meters, as we will be using the metric system in our calculations. 1 foot is approximately 0.3048 meters. So, 600 feet is equal to 182.88 meters.
Now, let's consider the horizontal and vertical components of the arrow's motion. Since the arrow is shot at an angle of 45 degrees, the initial velocity can be split into two components: one in the horizontal direction and another in the vertical direction.
The horizontal component of the initial velocity remains constant throughout the motion and is equal to the speed of the arrow when it leaves the bow. So, we need to find the value of the initial velocity in meters per second.
The vertical component of the initial velocity is affected by gravity. The arrow experiences an acceleration downwards due to gravity of 32 feet per second squared (9.8 meters per second squared). At the highest point of the arrow's trajectory, it momentarily stops moving vertically and starts moving downwards. The time taken for this to happen is the time it takes for the arrow to reach the peak.
Using the kinematic equation for vertical motion, we can find the time taken to reach the peak. The equation is given as:
0 = V0y - g * t
Where:
- V0y is the initial vertical velocity component
- g is the acceleration due to gravity (32 feet per second squared or 9.8 meters per second squared)
- t is the time taken to reach the peak
Since the initial vertical velocity component is the product of the initial velocity and the sine of the launch angle, we can rewrite the equation as follows:
0 = V0 * sin(45) - g * t
Simplifying, we get:
t = V0 * sin(45) / g
To find the time it takes for the arrow to hit the ground, we can double the value of the time taken to reach the peak as the total flight time of the arrow.
Now, we can use the equation for the horizontal component of motion:
range = V0 * cos(45) * t
We are given the range as 182.88 meters, the launch angle as 45 degrees, and acceleration due to gravity as 9.8 meters per second squared. We can substitute these values into the equations to find the initial velocity and calculate its value in miles per hour.
Let's solve the equations step by step:
1. Calculate the time taken to reach the peak:
t = V0 * sin(45) / g
t = V0 * (1/√2) / 9.8
2. Calculate the total flight time:
total_time = 2 * t
3. Calculate the initial velocity in the horizontal direction:
range = V0 * cos(45) * total_time
Solving for V0:
V0 = range / (cos(45) * total_time)
Finally, we can convert the initial velocity from meters per second to miles per hour:
1 mile = 1609.34 meters
1 hour = 3600 seconds
V0_mph = (V0 * 3600) / 1609.34
By performing the calculations, we will obtain the speed at which the arrow leaves the bow, measured in miles per hour.