What Launch Angle Does an Archer Fish Need to Hit Its Target?

[Swansong]

Homework Statement

Upon spotting an insect on a twig overhanging water, an archer fish squirts water drops at the insect to knock it into the water (Fig. 4 -38). Although the fish sees the insect along a straight-line path at angle f and distance d, a drop must be launched at a different angle θ if its parabolic path is to intersect the insect. If φ= 36.0° and d 0.900 m, what launch angle (theta) is required for the drop to be at the top of the parabolic path when it reaches the insect?
2. Relevant equation
So the two equations i have used so far are, y-yi=vi*sin(θ)-.5gt^2, x-xi=vi*cos(θ).

The Attempt at a Solution

Firstly, I found both xi and yi by using the trigonometric relationships. .9sin(φ)=.53 and .9cos(φ)=.73 for the vertical and horizontal distances. This yields the maximum height and distance from the initial position to the insect. I substituted .53 into the horizontal distance equation which gave me .73/(vi*cos(θ))= t, which i then substituted into the vertical distance equation. This is the part where I am stuck. After simplifying the equation, i am left with two unknown variables, vi and θ. 0= (-3.577*vi^2-.53)+tan(θ)-3.577vi^2*tan^2(θ). Should I solve the quadratic equation for θ and go from there?? p.s Sorry that I didn't use LaTeX, I'm not very comfortable with it, since I've never used it before, but i will try to use it in my following posts.

 

[Student100]

Homework Statement

Upon spotting an insect on a twig overhanging water, an archer fish squirts water drops at the insect to knock it into the water (Fig. 4 -38). Although the fish sees the insect along a straight-line path at angle f and distance d, a drop must be launched at a different angle θ if its parabolic path is to intersect the insect. If φ= 36.0° and d 0.900 m, what launch angle (theta) is required for the drop to be at the top of the parabolic path when it reaches the insect?
2. Relevant equation
So the two equations i have used so far are, y-yi=vi*sin(θ)-.5gt^2, x-xi=vi*cos(θ).

The Attempt at a Solution

Firstly, I found both xi and yi by using the trigonometric relationships. .9sin(φ)=.53 and .9cos(φ)=.73 for the vertical and horizontal distances. This yields the maximum height and distance from the initial position to the insect. I substituted .53 into the horizontal distance equation which gave me .73/(vi*cos(θ))= t, which i then substituted into the vertical distance equation. This is the part where I am stuck. After simplifying the equation, i am left with two unknown variables, vi and θ. 0= (-3.577*vi^2-.53)+tan(θ)-3.577vi^2*tan^2(θ). Should I solve the quadratic equation for θ and go from there?? p.s Sorry that I didn't use LaTeX, I'm not very comfortable with it, since I've never used it before, but i will try to use it in my following posts.

Try solving for time, then you should be able to solve for vi.

 

[Nathanael]

It's good practice to find the final answer in terms of the variables (D and Φ) and wait until the end to plug in the given values.

It may be easier to work with the initial components of velocity (V_x, V_y) instead of the speed/angle (V_i, θ). Quadratic equations can be avoided and the answer can be found as θ=arctan(V_y/V_x)

 

[Swansong]

It's good practice to find the final answer in terms of the variables (D and Φ) and wait until the end to plug in the given values.

It may be easier to work with the initial components of velocity (V_x, V_y) instead of the speed/angle (V_i, θ). Quadratic equations can be avoided and the answer can be found as θ=arctan(V_y/V_x)

How would i work with these initial components for this problem? wouldn't i still need to know initial velocity in this case?

 

[Nathanael]

How would i work with these initial components for this problem? wouldn't i still need to know initial velocity in this case?

You know the maximum height the water reaches, right? You can use this to determine the initial vertical velocity V_y.

In the time it takes the water to reach that height, you know how far horizontally it has moved. So then you can determine V_x that way.

 

[Swansong]

yes, I know the maximum height and range of the water. But I do not see how i can get the initial velocities from this. Are you referring to the equations, Xf=Vi*t*cos(θ) and Yf=Vi*t*sin(θ)-.5gt^2? Still kind of confused, sorry

 

[Nathanael]

Sorry I did not mean to be confusing, it just seemed like a simpler method.

V_y can be found from conservation of energy; or it can be found from the kinematics equation $y=V_yT-0.5gT^2$ where T is the time it takes to reach the top of it's path, (a.k.a. the time for V_y to decrease to zero, which is the T that satisfies the equation V_y-gT=0).

To find V_x you will use the same time T in the horizontal distance equation.

 

[Swansong]

Sorry I did not mean to be confusing, it just seemed like a simpler method.

V_y can be found from conservation of energy; or it can be found from the kinematics equation $y=V_yT-0.5gT^2$ where T is the time it takes to reach the top of it's path, (a.k.a. the time for V_y to decrease to zero, which is the T that satisfies the equation V_y-gT=0).

To find V_x you will use the same time T in the horizontal distance equation.

ohh, alright. Completely makes sense that way. Thank you for your help.