Upon spotting an insect on a twig overhanging water, an archer fish squirts water drops at the
insect to knock it into the water. Although the insect is at distance d from the fish along a
straight-line path at angle $ (as in the figure here), a drop must be launched at a different
angle if its parabolic path is to intersect the insect. If $ = 32.0° and d = 0.900 m, what is required for the drop to be at the top of the parabolic path when it reaches the insect?
To determine what is required for the drop to be at the top of the parabolic path when it reaches the insect, we need to consider the projectile motion of the drop and calculate the required initial velocity.
The initial velocity of the drop can be calculated using the components of velocity. The horizontal component of velocity remains constant throughout the motion, while the vertical component changes due to the acceleration due to gravity.
Given that the angle $ is 32.0° and the distance d is 0.900 m, we can break down the problem into horizontal and vertical components.
The horizontal component of velocity remains constant, so we can write:
Vx = initial velocity * cos($)
The vertical component of velocity changes due to the acceleration due to gravity. At the highest point of the parabolic path, the vertical component of velocity becomes zero. We can find the time t it takes for the drop to reach the highest point using the vertical motion equation:
0 = initial vertical velocity + (-9.8 m/s^2) * t
The initial vertical velocity can be written as:
Vy = initial velocity * sin($)
Solving the equation for the time t:
0 = initial velocity * sin($) + (-9.8 m/s^2) * t
t = initial velocity * sin($) / 9.8 m/s^2
The total time of flight can be found by doubling the time it takes to reach the highest point:
T = 2 * t
Now, we can find the horizontal displacement of the drop during this time t:
d = Vx * T
Substituting the expressions for Vx and T:
d = initial velocity * cos($) * (2 * (initial velocity * sin($) / 9.8 m/s^2))
Simplifying the equation:
d = (2 * initial velocity^2 * (sin($) * cos($))) / 9.8 m/s^2
Now we can solve for the initial velocity:
initial velocity = √((d * 9.8 m/s^2) / (2 * (sin($) * cos($))))
Plug in the given values for $ and d:
initial velocity = √((0.900 m * 9.8 m/s^2) / (2 * (sin(32.0°) * cos(32.0°))))
Calculating the equation:
initial velocity ≈ 2.13 m/s
Therefore, the required initial velocity for the drop to be at the top of the parabolic path when it reaches the insect is approximately 2.13 m/s.