The small archerfish (length 20 to 25 cm) lives in brackish waters of Southeast Asia from India to the Philippines. This aptly named creature captures its prey by shooting a stream of water drops at an insect, either flying or at rest. The bug falls into the water and the fish gobbles it up. The archerfish has high accuracy at distances of 1.2 m to 1.5 m, and it sometimes makes hits at distances up to 3.5 m. A groove in the roof of its mouth, along with a curled tongue, forms a tube that enables the fish to impart high velocity to the water in its mouth when it suddenly closes its gill flaps. Suppose the archerfish shoots at a target that is 2.10 m away, measured along a line at an angle of 30.0° above the horizontal. With what velocity must the water stream be launched if it is not to drop more than 4.00 cm vertically on its path to the target?
i have no clue on how to start this problem please help :/ i've attempted three times already
oh wow :o i just figured out the problem but your way was much more simple then mines. ty :)
To solve this problem, we can apply the principles of projectile motion. Here are the steps you can follow to find the velocity required for the water stream:
Step 1: Analyze the problem
In this case, we have a projectile motion problem involving the motion of a stream of water. We are given the distance to the target, the angle above the horizontal, and the maximum allowable vertical drop. We need to find the required velocity.
Step 2: Break down the initial velocity
We can express the initial velocity of the water stream as a sum of its horizontal and vertical components. Let's assume the initial velocity of the water stream is v, and the angle above the horizontal is θ.
The horizontal component of velocity, Vx, can be found using:
Vx = v * cos(θ)
The vertical component of velocity, Vy, can be found using:
Vy = v * sin(θ)
Step 3: Analyze the vertical motion
In projectile motion, the vertical motion can be analyzed independently using the equations of motion. Since we are given the maximum allowable drop of 4.00 cm, we need to find the time it takes for the water stream to reach the target.
We can use the equation for vertical displacement:
Δy = Vy * t + (1/2) * g * t^2
Since the water stream should not drop more than 4.00 cm vertically, we have:
(0.04 m) ≤ Vy * t + (1/2) * (9.8 m/s^2) * t^2
Step 4: Solve the equation for time
Rearrange the equation to a quadratic form:
(1/2) * (9.8 m/s^2) * t^2 + Vy * t - 0.04 m = 0
Solve this quadratic equation for t using any suitable method (e.g., quadratic formula or factoring).
Step 5: Determine the required velocity
Now that you have found the time it takes for the water stream to reach the target, you can use this information to determine the required velocity. Recall that Vy = v * sin(θ).
Using the found value of t, substitute it back into the equation Δy = Vy * t + (1/2) * g * t^2 to find the vertically dropped distance. Calculate the difference between this and the maximum allowable drop (4.00 cm).
Next, solve for the magnitude of the initial velocity, v, by using the equation:
v = (Δy + 0.04 m) / (sin(θ) * t)
Step 6: Calculate the horizontal velocity component
Finally, substitute the found value of v into Vx = v * cos(θ) to find the horizontal component of velocity, Vx.
This calculated value of Vx is the velocity at which the water stream should be launched to hit the target without dropping more than 4.00 cm vertically.
Make sure to double-check the calculations and units to ensure accuracy.
No worries, I'm here to help you. Let's break down the problem step by step.
First, we need to determine the initial velocity of the water stream needed for it to reach the target without dropping more than 4.00 cm.
To do this, we can use the equations of motion and consider the vertical and horizontal components separately.
Vertical Component:
The water stream will follow a projectile motion due to gravity. We can use the following equation to calculate the vertical displacement (Δy) of the water stream:
Δy = v₀y * t + (1/2) * g * t²
Here, v₀y is the initial vertical velocity, t is the time of flight, and g is the acceleration due to gravity (approximately 9.8 m/s²).
In this case, we want the vertical displacement to be less than or equal to 4.00 cm, which is 0.04 m.
Horizontal Component:
The horizontal displacement of the water stream is given by:
Δx = v₀x * t
Here, v₀x is the initial horizontal velocity.
We are given the horizontal distance (Δx) as 2.10 m and the angle (θ) as 30.0°. Thus, we can find the initial horizontal velocity using the following equation:
v₀x = Δx / (t * cos(θ))
Now, we can determine the time of flight by using the formula for the horizontal displacement:
Δx = v₀ * cos(θ) * t
Rearranging the equation gives us:
t = Δx / (v₀ * cos(θ))
Now, let's substitute the values into the equations and solve step by step.
1. Convert the 4.00 cm vertical displacement to meters:
Δy = 0.04 m
2. Calculate the initial horizontal velocity:
v₀x = Δx / (t * cos(θ))
v₀x = 2.10 m / (t * cos(30.0°))
3. Find the time of flight:
Δx = v₀ * cos(θ) * t
2.10 m = v₀ * cos(30.0°) * t
t = 2.10 m / (v₀ * cos(30.0°))
4. Substitute the value for t in the vertical component equation:
Δy = v₀y * t + (1/2) * g * t²
0.04 m = v₀y * [2.10 m / (v₀ * cos(30.0°))] + (1/2) * g * [2.10 m / (v₀ * cos(30.0°))]²
Now, we have an equation in terms of v₀ (the initial velocity) to solve for. Let me know if you would like me to continue with the calculations.
drop= .04=1/2 g t^2 solve for time to drop 4cm.
velocity=2.10/timeabove