# Solving for Angle to Destroy Rocket Launcher

• Idontgetphysics
In summary: Cos) (Theta)] Squared.The 141 is the muzzle velocity of the patrol boat's stabilized 30-mm cannon. The 1500 is the distance between the patrol boat and the rocket launcher. The (1500/141) is the muzzle velocity divided by the distance. The 4.9 is the muzzle velocity squared. The (1500/141) is the muzzle velocity multiplied by the distance divided by the muzzle velocity.
Idontgetphysics

## Homework Statement

A patrol boat off a coast is ordered to destroy a rocket launcher that sits on the roof of 40-meter tall building. The horizontal distance between the boat and the rocket launcher is 1500 meters. The muzzle velocity of the patrol boat’s stabilized 30-mm cannon is 141 meters/second. What are the two angles (relative to the horizontal) at which the patrol boat’s gun can be aimed in order to destroy the rocket launcher? Express your answer in degrees to a precision of three significant figures. [You may ignore air resistance. Assume an acceleration of gravity of precisely 9.800 m/s2. Your answers must be accurate to within 0.3 degrees in order to receive credit. You must also show your work.]
Vo= 141 m/s
X= 1500 m
Yf = 40 m
Yo = 0
t = ?
Theta = ?

## Homework Equations

Yf=Yo + Voy(t) - .5(9.8)(tsquared)

## The Attempt at a Solution

One angle seems to be 23.4 degrees but I can't understand how you would find the other angle. I divided the initial velocity by the x distance to find 10.34 seconds. Plugged that into the Yf=Yo + Voy(t) - .5(9.8)(tsquared) equation with Yf being 40 meters, Yo being zero and Voy being (Vo)(sin)(theta) and then solving for theta. I checked it and it is accurate within a meter to the final position of y being 40 meters so the math is accurate even if my theory may be incorrect. However, I don't see any strategy for finding the second angle.

Last edited by a moderator:
Idontgetphysics said:
divided the initial velocity by the x distance to find 10.34
I assume you mean you divided the horizontal distance by the velocity, but that is not the horizontal velocity.

Your one equation has two unknowns: v0y and t. That will not be enough. You need another equation. You should probably consider the motion in x.

Anyone have any ideas on what to do here?

Idontgetphysics said:
Anyone have any ideas on what to do here?
Let the angle be θ. What is the horizontal velocity? What is the time to the target?

haruspex said:
Let the angle be θ. What is the horizontal velocity? What is the time to the target?
Yes, I am aware that the angle is theta. There is no other information provided, You must deduce the angle from ONLY those three variables - Vo is 141 m/s. Final y position I assume is 40 meters and the x distance is 1500 meters. I've taken the derivative for time to be t = X / (Vo) (Cos) (Theta) and then solving for theta which I've worked is the logical means to the solution. I plug my derivative for T into the equation Yf = Yi + gt - 4.9 t squared but unfortunately my math skill are too broken to be able to work the math out sufficiently. I've attempted three times and checked it against the equation and it does not work. Basically, I am just asking someone to confirm my theory and solve the math for my equation which becomes, with all variables, 40 = 0 + (141) [( 1500/ (141) (Cos) (Theta)] - 4.9 ([( 1500/ (141) (Cos) (Theta)] Squared and then solving for Theta. My math skills are not good enough to solve this equation.

There's a square root in there, so you need to consider both the positive and negative roots.

You are very close. Using the x equation to get t in terms of v0 and cos(theta) and substituting into the y equation is one way to do it. You might find it easier to solve the y equation for t and substitute it into x. As you correctly noted, the rest is just solving the equation. We’ll talk about that in a second.

However, first, I do see one mistake. You wrote

Yf = Yi + gt - 4.9 t squared

Why g t? g isn’t the initial velocity in the y dimension (hint hint)

Now, about solving the equation. You will find you have a result which includes both sines and cosines, and which can’t be easily solved for theta. Note that your problem statement indicates that the answer must be within 0.3 degrees. This is a good indication that the problem requires approximation or a numerical solution. You can either approximate the trig functions, or find the two solutions numerically

Idontgetphysics said:
40 = 0 + (141) [( 1500/ (141) (Cos) (Theta)] - 4.9 ([( 1500/ (141) (Cos) (Theta)] Squared
Not quite. Please explain all the terms in this part: (141) [( 1500/ (141) (Cos) (Theta)]

## 1. What is "Solving for Angle to Destroy Rocket Launcher"?

"Solving for Angle to Destroy Rocket Launcher" refers to the process of determining the angle at which a projectile must be launched in order to successfully hit and destroy a stationary or moving rocket launcher.

## 2. Why is it important to solve for the angle to destroy a rocket launcher?

It is important to solve for the angle to destroy a rocket launcher because it allows for precise and accurate targeting, which is crucial in situations where lives and resources are at stake. Additionally, understanding the angle required for destruction can also aid in improving weapon effectiveness and efficiency.

## 3. What factors are involved in solving for the angle to destroy a rocket launcher?

The factors involved in solving for the angle to destroy a rocket launcher include the initial velocity of the projectile, the distance between the launcher and the target, the gravitational force, and the air resistance. Other factors such as wind speed and direction may also need to be considered.

## 4. How do scientists solve for the angle to destroy a rocket launcher?

Scientists use mathematical equations and principles, such as Newton's laws of motion and projectile motion equations, to solve for the angle to destroy a rocket launcher. They also take into account various factors such as the velocity and trajectory of the projectile, as well as external forces like gravity and air resistance.

## 5. Are there any practical applications for solving for the angle to destroy a rocket launcher?

Yes, there are many practical applications for solving for the angle to destroy a rocket launcher. It is used in military operations to accurately target and destroy enemy weapons, as well as in sports such as archery and long-range shooting. It can also be applied in engineering and design to improve the effectiveness of projectile weapons.

• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
4K
• Introductory Physics Homework Help
Replies
6
Views
6K
• Introductory Physics Homework Help
Replies
4
Views
5K
• Introductory Physics Homework Help
Replies
4
Views
3K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
3K