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 1.75 m away, 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 1.60 cm vertically on its path to the target?
To solve this problem, we can use the principles of projectile motion. Let's break down the given information and solve step by step:
Given:
- Distance to the target (horizontal displacement): 1.75 m
- Angle above the horizontal: 30.0°
- Vertical displacement (drop allowed): 1.60 cm
To find the launch velocity of the water stream, we need to calculate its horizontal and vertical components. First, let's find the time of flight (t) for the water stream.
Step 1: Calculate time of flight (t)
We know that the horizontal displacement is equal to the horizontal component of velocity (Vx) multiplied by time (t).
Using the equation: horizontal displacement = Vx * t
1.75 m = Vx * t ---(Equation 1)
Next, let's find the vertical displacement (Δy) of the water stream using the given information.
Step 2: Calculate vertical displacement (Δy)
We know that the vertical displacement is equal to the vertical component of velocity (Vy) multiplied by the time (t) plus half the acceleration due to gravity (g) multiplied by the square of the time (t).
Using the equation: vertical displacement = Vy * t + (1/2) * g * t^2
1.60 cm = Vy * t + (1/2) * (-9.8 m/s^2) * t^2 ---(Equation 2)
Note: 1 cm = 0.01 m, so 1.60 cm = 0.016 m
Step 3: Calculate Vy
Given that the angle above the horizontal (θ) is 30.0°, we can find Vy by using the equation:
Vy = V * sin(θ), where V is the magnitude of the launch velocity.
Step 4: Substitute values into the equations
Substitute the value of Vy in Equations 1 and 2, and solve the system of equations to find Vx and V.
1. Solve Equation 2 for Vy:
0.016 m = Vy * t + (1/2) * (-9.8 m/s^2) * t^2
0.016 m = (V * sin(30°)) * t - 4.9 m/s^2 * t^2 ---(Equation 3)
2. Substitute Vy from Equation 3 into Equation 1:
1.75 m = (V * cos(30°)) * t ---(Equation 4)
Step 5: Solve the system of equations
Now we have two equations (Equations 3 and 4) with two unknowns (V and t). We'll solve these equations simultaneously.
Divide Equation 3 by 4.9 to simplify:
(0.016 m / 4.9 m/s^2) = [(V * sin(30°)) / 4.9 m/s^2] * t - (t^2)
0.00327 s = (0.203 * V) * t - (t^2) ---(Equation 5)
Substitute V * cos(30°) from Equation 4 into Equation 5:
0.00327 s = (0.203 * V * cos(30°)) * t - (t^2)
Rearrange into quadratic form:
0.00327 t^2 - (0.203 * V * cos(30°)) * t + 0.00327 = 0
Now, this is a quadratic equation in the form of: at^2 + bt + c = 0, where:
a = 0.00327
b = -(0.203 * V * cos(30°))
c = 0.00327
Step 6: Solve the quadratic equation
Use the quadratic formula to find t:
t = (-b ± √(b^2 - 4ac)) / (2a)
Substitute the values and solve for t.
Step 7: Calculate V
Substitute the value of t back into Equation 4 to calculate V.
V = (1.75 m) / [(cos(30°)) * t]
Solve for V.
After following these steps, you will find the launch velocity (V) required for the water stream to not drop more than 1.60 cm vertically on its path to the target.
To solve this problem, we can use the principles of projectile motion. Let's break down the problem and solve it step-by-step:
Step 1: Determine the horizontal and vertical components of the initial velocity of the water stream.
The given launch angle is 30.0° above the horizontal. We can decompose the initial velocity into horizontal and vertical components:
V₀x = V₀ * cosθ
V₀y = V₀ * sinθ
Where:
V₀x is the horizontal component of the initial velocity,
V₀y is the vertical component of the initial velocity,
V₀ is the magnitude of the initial velocity, and
θ is the launch angle of 30.0°.
Step 2: Calculate the time it takes for the water stream to travel 1.75 m horizontally.
Since the vertical displacement of the water stream is given as 1.60 cm (0.016 m), it won't affect the horizontal motion. Therefore, we can focus on the horizontal displacement.
The horizontal distance traveled (range) is given as 1.75 m. We can use the horizontal velocity component and the time of flight to determine the time it takes for the water stream to reach the target.
Range = V₀x * t
Solving for time (t):
t = Range / V₀x
Step 3: Calculate the time of flight.
The total time of flight is the time taken for the water stream to reach maximum height and then fall back to the same vertical position. Since the initial and final vertical positions are the same, the time taken to reach the maximum height is half of the total time of flight.
The time of flight can be determined using the vertical component of the initial velocity and the acceleration due to gravity (g = 9.8 m/s²) considering the vertical motion.
Δy = V₀y * t + (1/2) * g * t²
Since Δy = 0 (the water stream starts and ends at the same elevation), we can solve for t:
0 = V₀y * t + (1/2) * g * t²
Simplifying the equation:
(1/2) * g * t² = -V₀y * t
Divide both sides by t and rearrange the equation:
(1/2) * g * t = -V₀y
Solving for t:
t = -2 * V₀y / g
Step 4: Substitute the values into the equations and solve for V₀.
Using the information we obtained in steps 2 and 3, we can substitute the values to find V₀:
t = Range / V₀x
t = -2 * V₀y / g
Substituting V₀x = V₀ * cosθ and V₀y = V₀ * sinθ:
Range / (V₀ * cosθ) = -2 * (V₀ * sinθ) / g
Simplifying the equation:
Range * g = -2 * sinθ * cosθ * V₀
Finally, solving for V₀:
V₀ = -(Range * g) / (2 * sinθ * cosθ)
Substituting the given values, the unit of velocity should be in m/s.
Calculating V₀:
V₀ = -[(1.75 m) * (9.8 m/s²)] / [2 * sin(30.0°) * cos(30.0°)]