As discussed in Example 4-7, the archerfish hunts by dislodging an unsuspecting insect from its resting place with a stream of water expelled from the fish's mouth. Suppose the archerfish squirts water with a speed of 1.4 m/s at an angle of 50.0° above the horizontal, and aims for a beetle on a leaf 3.0 cm above the water's surface.
(a) At what horizontal distance from the beetle should the archerfish fire if it is to hit its target in the least time?
(b) How much time will the beetle have to react?
To determine the horizontal distance the archerfish should fire from to hit its target in the least time, we can use the equations of motion and projectile motion.
(a) Let's break down the problem into two dimensions: the horizontal (x) and vertical (y) directions.
In the horizontal direction (x), the velocity is constant because there is no horizontal acceleration. Therefore, the horizontal distance traveled by the water stream is given by:
x = vx * t
In the vertical direction (y), the water stream experiences free-fall motion with uniform acceleration due to gravity. The vertical distance traveled by the water stream is given by:
y = vy0 * t - (1/2) * g * t^2
In this case, the initial vertical velocity vy0 is calculated from the initial speed of the water stream expelling from the fish's mouth and the launch angle:
vy0 = v0 * sin(θ)
The time it takes for the water stream to reach the beetle's height (3.0 cm) above the water's surface is the same time it takes for the beetle to react (assuming it reacts instantly). Thus, we can solve for t by setting the vertical distance (y) equal to 3.0 cm:
3.0 cm = vy0 * t - (1/2) * g * t^2
Next, we need to express all the measurements in the appropriate units. In this case, we will convert the vertical distance to meters:
3.0 cm = 0.03 m
Plugging in the values and solving the quadratic equation will give us the time it takes for the beetle to react (t).
Once we have the time (t), we can substitute it back into the horizontal equation (x = vx * t) to find the horizontal distance at which the archerfish should fire to hit its target in the least time.
(b) To calculate the time the beetle has to react, we solve the quadratic equation obtained from the vertical distance equation in part (a).
Let's go ahead and calculate the answers.
To solve this problem, we can use the principles of projectile motion. Let's break it down step-by-step:
Step 1: Define the given variables
- The water squirting speed (v) = 1.4 m/s
- The angle above the horizontal (θ) = 50.0°
- The vertical displacement of the beetle (y) = 3.0 cm above the water's surface
Step 2: Convert the vertical displacement to meters
- 3.0 cm = 0.03 m (1 cm = 0.01 m)
Step 3: Determine the horizontal distance (x) from the beetle
- Since the water is being squirted at an angle above the horizontal, we need to split the velocity into horizontal and vertical components.
- The horizontal component of the velocity (v_x) can be found using the following equation: v_x = v*cos(θ)
- The time of flight (t) can be determined using the equation: t = 2*(v*sin(θ))/g (where g is the acceleration due to gravity, approximately 9.8 m/s^2)
- The horizontal distance (x) can be found using the equation: x = v_x*t
Step 4: Calculate the horizontal distance (x)
- Substitute the given values into the equations:
v_x = 1.4 m/s * cos(50.0°)
t = 2 * (1.4 m/s * sin(50.0°)) / 9.8 m/s^2
x = v_x * t
Step 5: Calculate the time for the beetle to react
- The time for the beetle to react is equal to the time of flight (t)
Let's calculate the results:
Step 4:
v_x = 1.4 m/s * cos(50.0°) ≈ 0.901 m/s
t = 2 * (1.4 m/s * sin(50.0°)) / 9.8 m/s^2 ≈ 0.332 s
x = v_x * t
x = 0.901 m/s * 0.332 s ≈ 0.299 m (or 29.9 cm)
Therefore, the archerfish should fire at a horizontal distance of approximately 0.299 m (or 29.9 cm) from the beetle.
Step 5:
The beetle will have to react for approximately 0.332 seconds.