# Solving for Launch Angle Using Vo

• Alex T Weliever
In summary, the problem asks to find the angle at which a projectile is launched, given an initial net velocity of 430 m/s and the projectile landing on the edge of an 80 meter high cliff 175 meters away. The relevant equations are tan^2+1=sec^2 and the basic projectile motion equations for the vertical and horizontal components. To solve for the angle, eliminate time from the vertical motion equation using the horizontal equation.
Alex T Weliever

## Homework Statement

Find the angle at which the projectile is launched.
(Hint: use tan^2+1=sec^2)

Initial Net Velocity=430 m/s
The projectile lands perfectly on the edge of a cliff 80 meters high and 175 meters away.

## Homework Equations

tan^2+1=sec^2
Kinematics, i.e.-
v=vo+gt
dy=vo(dt)+.5(-10)dt^2
v^2=Vo^2+2(-10)dy
and so on.

## The Attempt at a Solution

I've gotten as far as listing my knowns and solving out for a few things implicitly, nothing concrete really solved.

Vox=430cos theta
Voy=430 sin theta
I believe the following is true?
43*sin theta=t(peak)
86*sin theta=t(final, not taking the cliff ledge into account)

What I'm struggling with is just beginning, really. I don't want someone to tell me the answer, just guide me to what I need to start with. I started by focusing on the Vox/Voy triangle to solve for velocities, but then I realized with that information I could solve for the angle automatically, and that wouldn't involve the ledge part, so I can't do that. I realize (assuming this is correct) that I can't use 175 as dx or 80 as dy as the projectile would go farther and lower if the cliff ledge was not there, so I'm unsure of how to solve for that. Looking for steps in the right direction. Thanks!

Hi Alex, Welcome to Physics Forums!

Start with the basic projectile motion equations for the vertical and horizontal components. Eliminate time from the vertical motion equation using the horizontal equation. You should be able to see the way forward from there.

## 1. What is the formula for calculating launch angle using initial velocity (Vo)?

The formula for calculating launch angle using initial velocity is: θ = arcsin (g * t / Vo), where θ is the launch angle, g is the acceleration due to gravity (9.8 m/s²), t is the time in seconds, and Vo is the initial velocity in meters per second.

## 2. How is solving for launch angle using Vo useful in scientific research?

Solving for launch angle using Vo is useful in scientific research because it allows researchers to accurately predict the trajectory of an object in motion, such as a projectile, and understand how different variables, such as initial velocity and launch angle, affect its flight path.

## 3. Can launch angle be calculated using any initial velocity?

Yes, launch angle can be calculated using any initial velocity. However, the object's launch angle will vary depending on the initial velocity and other factors, such as air resistance and gravity.

## 4. How does changing the launch angle affect the trajectory of an object?

Changing the launch angle can significantly affect the trajectory of an object. A lower launch angle will result in a shorter and lower flight path, while a higher launch angle will result in a longer and higher flight path. The launch angle also determines the angle at which the object will hit the ground or a target.

## 5. Are there any limitations to using Vo to solve for launch angle?

Yes, there are some limitations to using Vo to solve for launch angle. This formula assumes a perfect projectile motion, which may not always be the case in real-world scenarios. Other factors such as air resistance, wind, and surface conditions can also affect the actual trajectory of an object.

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