An 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 2.1 m/s at an angle of 51 degrees above the horizontal, aiming for a beetle on a leaf that is 3 cm above the water’s surface. At what horizontal distance (in cm) should the archerfish be from the beetle to minimize the time the beetle has to react?
To find the horizontal distance from the beetle to the archerfish that minimizes the time the beetle has to react, we need to calculate the time it takes for the water stream to travel from the fish's mouth to the beetle.
First, we convert the height of the beetle to meters:
3 cm = 3/100 m = 0.03 m
The initial vertical velocity of the water stream is given by:
Viy = speed * sin(angle)
Viy = 2.1 m/s * sin(51 degrees) ≈ 1.63 m/s
The time it takes for the water stream to reach maximum height (the same level as the beetle) can be calculated using the equation:
Vfy = Viy - g*t
where Vfy is the final vertical velocity (which is zero when the water stream reaches maximum height), Viy is the initial vertical velocity, g is the acceleration due to gravity (9.8 m/s^2), and t is time.
Plugging in the values, we get:
0 = 1.63 m/s - 9.8 m/s^2 * t
Solving for t:
t = 1.63 m/s / 9.8 m/s^2 ≈ 0.17 s
The time it takes for the water stream to reach the beetle can be found by calculating the total time it takes for the water stream to hit the beetle's height:
t_total = 2 * t
t_total = 2 * 0.17 s = 0.34 s
To calculate the horizontal distance (x) traveled by the water stream, we use the equation:
x = speed * cos(angle) * t_total
Plugging in the values, we get:
x = 2.1 m/s * cos(51 degrees) * 0.34 s
Finally, we convert the distance to centimeters:
x_cm = x * 100 cm/m
Now we can calculate the result:
x = 2.1 m/s * cos(51 degrees) * 0.34 s * 100 cm/1 m = approximately 110.5 cm
Therefore, the archerfish should be approximately 110.5 cm away from the beetle to minimize the time the beetle has to react.
To solve the problem, we need to find the horizontal distance traveled by the water stream from the archerfish's mouth to the target beetle. This will help us determine the shooting distance or the distance at which the archerfish should be from the beetle to minimize the reaction time.
To calculate the horizontal distance, we can break the initial velocity of the water stream into its horizontal and vertical components. The horizontal component represents the velocity in the direction parallel to the water's surface, while the vertical component represents the velocity in the upward direction.
Given:
- Water stream speed (initial velocity) = 2.1 m/s
- Angle above the horizontal = 51 degrees
- Beetle's height above the water's surface = 3 cm
We need to convert the beetle's height to meters for consistent units. 1 cm = 0.01 m, so the height of the beetle above the water's surface is:
Height = 3 cm * 0.01 m/cm = 0.03 m
Now, let's find the components of velocity:
Horizontal component = initial velocity * cos(angle)
Vertical component = initial velocity * sin(angle)
Horizontal component = 2.1 m/s * cos(51 degrees)
Vertical component = 2.1 m/s * sin(51 degrees)
Next, we can calculate the time the water stream takes to reach the beetle at a vertical displacement of 0.03 m:
Time = Vertical displacement / Vertical component
Time = 0.03 m / Vertical component
Finally, to find the horizontal distance traveled by the water stream:
Horizontal distance = Horizontal component * Time
Horizontal distance = Horizontal component * (0.03 m / Vertical component)
Now, we can substitute the values into the equation to calculate the horizontal distance.