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 2.15 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 3.50 cm vertically on its path to the target?
To solve this problem, we need to find the initial velocity of the water stream launched by the archerfish. We can break this motion into horizontal and vertical components.
First, let's find the horizontal component of the initial velocity. The horizontal distance traveled by the water stream is given as 2.15 meters. The time it takes for the water stream to reach the target horizontally is the same as the time taken by a projectile with the same horizontal range and initial velocity. We can use the formula for horizontal distance traveled:
Horizontal distance = (initial horizontal velocity) * time
The initial vertical velocity of the water stream is zero because it starts from rest vertically. Therefore, the time taken for the water stream to reach the target horizontally will be the same as the time taken for it to fall vertically.
Next, let's find the vertical component of the initial velocity. We know that the water stream should not drop more than 3.50 cm vertically on its path to the target. This means that the vertical distance traveled should be less than or equal to 3.50 cm. We can use the formula for vertical distance traveled:
Vertical distance = (initial vertical velocity) * time + (1/2) * acceleration * time^2
Since the water stream starts from rest vertically, the first term on the right side of the equation becomes zero. We can now solve for the initial vertical velocity.
Now we can calculate the initial velocity of the water stream by combining the horizontal and vertical components using the Pythagorean theorem:
Initial velocity = sqrt((initial horizontal velocity)^2 + (initial vertical velocity)^2)
Let's plug in the given values and solve for the initial velocity:
Horizontal distance = 2.15 m
Vertical distance = 3.50 cm = 0.035 m
Angle above horizontal = 30.0°
First, let's find the time taken for the water stream to reach the target horizontally:
Horizontal distance = (initial horizontal velocity) * time
time = Horizontal distance / (initial horizontal velocity)
time = 2.15 m / (initial horizontal velocity)
Next, let's find the initial vertical velocity:
Vertical distance = (initial vertical velocity) * time + (1/2) * acceleration * time^2
0.035 m = (initial vertical velocity) * [2.15 m / (initial horizontal velocity)] + (1/2) * (-9.8 m/s^2) * [2.15 m / (initial horizontal velocity)]^2
Now, let's solve for the initial vertical velocity:
0.035 m = (initial vertical velocity) * [2.15 m / (initial horizontal velocity)] - 9.8 m/s^2 * [2.15 m / (initial horizontal velocity)]^2 / 2
(initial vertical velocity) = (0.035 m + 9.8 m/s^2 * [2.15 m / (initial horizontal velocity)]^2 / 2) / [2.15 m / (initial horizontal velocity)]
Finally, let's calculate the initial velocity using the Pythagorean theorem:
Initial velocity = sqrt((initial horizontal velocity)^2 + (initial vertical velocity)^2)
By substituting the values we obtained, we can calculate the initial velocity of the water stream launched by the archerfish.