# Angled Projectile at an elevated height w/o a given angle

• darkdeeds
In summary, the conversation discussed the topic of game physics in modern 3D video games and how it relates to designing realistic movements for objects. The conversation then focused on a specific scenario in a game where a cannon must be fired at a target on a distant cliff. The target is at a distance of 400m and a slight elevation of 150m above the player's position. The launch speed of the cannonball is 100m/s. The conversation included equations and calculations to determine the necessary angles and time difference for the cannon to hit the target, taking into account the acceleration due to gravity.

## Homework Statement

Many modern 3D video games must be designed such that objects move in ways as close as possible to reality. This is often called “game physics”, which is run by a physics engine coded into the game. In one event of a recent game, a player’s job is to shoot a cannon at objects on a distant cliff. The target is supposed to be at a distance of 400 m and is at a slight elevation with respect to the player – say 150 m – and the launch speed of the cannonball is supposed to be 100 m/s relative to the player's “stationary” avatar. The game physics ignores wind resistance (many if not all of them do). At what angles should the cannon be fired in order to hit the target?

Enter the larger angle in degrees.

What is the smaller angle?

What is the time difference between the two trajectories?

ay=-9.81 m/s^2
dx=400 m
dy=150 m
Vo= 100 m/s
Angle?
time?

Vf=Vo+at
d=Vot+(1/2)at^2
d=(Vf^2-Vo^2)/2a

## The Attempt at a Solution

I found the greater angle to be 77.6 degrees. And by experience usually the smaller angle is simply 90-whatever the angle is to find the other. The smaller angle measures to be 12.4 degrees but this can't be right due tot he fact that the angle of elevation of the object is 20.55 degrees so the angle must be at least that. I'm not sure where to progress.

I made up a general formula for finding the distance (found by v=d/t)

d=(10000sin(2angle))/9.81

Help!

darkdeeds said:

## Homework Statement

Many modern 3D video games must be designed such that objects move in ways as close as possible to reality. This is often called “game physics”, which is run by a physics engine coded into the game. In one event of a recent game, a player’s job is to shoot a cannon at objects on a distant cliff. The target is supposed to be at a distance of 400 m and is at a slight elevation with respect to the player – say 150 m – and the launch speed of the cannonball is supposed to be 100 m/s relative to the player's “stationary” avatar. The game physics ignores wind resistance (many if not all of them do). At what angles should the cannon be fired in order to hit the target?

Enter the larger angle in degrees.
What is the smaller angle?
What is the time difference between the two trajectories?

ay=-9.81 m/s^2
dx=400 m
dy=150 m
Vo= 100 m/s
Angle?
time?

Vf=Vo+at
d=Vot+(1/2)at^2
d=(Vf^2-Vo^2)/2a

## The Attempt at a Solution

I found the greater angle to be 77.6 degrees. And by experience usually the smaller angle is simply 90-whatever the angle is to find the other. The smaller angle measures to be 12.4 degrees but this can't be right due tot he fact that the angle of elevation of the object is 20.55 degrees so the angle must be at least that. I'm not sure where to progress.

I made up a general formula for finding the distance (found by v=d/t)

d=(10000sin(2angle))/9.81

Help!
Using 90° - Angle1 = Angle2 only works if the target is at the same elevation as the launcher (cannon).

How did you find the first angle?

You gave few details as to how you arrived at your answer.

Hint: The time it takes for the projectile to reach the target is the same as each of the following:

The time it takes an object to travel 400 m at a constant velocity equal to the horizontal component of the velocity of your projectile.

The time it takes an object tossed straight up to return to an elevation of 150 meters if it is tossed upward with an initial velocity equal to the vertical component of the velocity of your projectile.

Last edited:
SammyS said:
Using 90° - Angle1 = Angle2 only works if the target is at the same elevation as the launcher (cannon).

How did you find the first angle?

You gave few details as to how you arrived at your answer.

Hint: The time it takes for the projectile to reach the target is the same as each of the following:

The time it takes an object to travel 400 m at a constant velocity equal to the horizontal component of the velocity of your projectile.

The time it takes an object tossed straight up to return to an elevation of 150 meters if it is tossed upward with an initial velocity equal to the vertical component of the velocity of your projectile.

I found my answer by putting 150^2+400^2=d^2 so that I could put d into my equation that i showed earlier.I came out with a value for theta, and it was 12.4, and the question asked for the larger angle so i substracted it from 90 degrees to get 77.6

darkdeeds said:
I found my answer by putting 150^2+400^2=d^2 so that I could put d into my equation that i showed earlier.I came out with a value for theta, and it was 12.4, and the question asked for the larger angle so i subtracted it from 90 degrees to get 77.6
12.4° can't be right. The line of sight, from the cannon to the target makes an angle of about 20.56° above the horizontal.

This is one of the harder type of projectile problem to do --- outside of those with air resistance.

Write out the kinematic equations for horizontal motion. This is constant velocity, so these are relatively easy.

Write out the kinematic equations for vertical motion. This is constant acceleration, not so simple as the above.

When I combine these I get either a quadratic equation in tan(θ) or a quadratic equation in t2 . Both look quite messy, although I left them in terms of parameters & variables, without substituting many numbers.

SammyS said:
This is one of the harder type of projectile problem to do --- outside of those with air resistance.

Write out the kinematic equations for horizontal motion. This is constant velocity, so these are relatively easy.

Write out the kinematic equations for vertical motion. This is constant acceleration, not so simple as the above.

When I combine these I get either a quadratic equation in tan(θ) or a quadratic equation in t2 . Both look quite messy, although I left them in terms of parameters & variables, without substituting many numbers.

When setting up the quadratic formula for tan(θ), do you let the vertical component equal 150?

darkdeeds said:
When setting up the quadratic formula for tan(θ), do you let the vertical component equal 150?
The vertical component of what?

## 1. What is an angled projectile at an elevated height without a given angle?

An angled projectile at an elevated height without a given angle refers to an object that is launched at an angle from an elevated position without knowing the exact angle at which it was launched.

## 2. How is the trajectory of an angled projectile at an elevated height without a given angle calculated?

The trajectory of an angled projectile at an elevated height without a given angle can be calculated using the equations of motion, specifically the equations for projectile motion. This requires knowing the initial velocity, time of flight, and the height of the object at different points along its path.

## 3. What factors affect the trajectory of an angled projectile at an elevated height without a given angle?

The factors that affect the trajectory of an angled projectile at an elevated height without a given angle include the initial velocity, the acceleration due to gravity, and the height of the launch point. Other factors such as air resistance and wind can also impact the trajectory.

## 4. Can the trajectory of an angled projectile at an elevated height without a given angle be predicted accurately?

In theory, the trajectory of an angled projectile at an elevated height without a given angle can be predicted accurately using the equations of motion. However, in reality, factors such as air resistance and wind can affect the trajectory and make it difficult to predict with complete accuracy.

## 5. How is the maximum height of an angled projectile at an elevated height without a given angle determined?

The maximum height of an angled projectile at an elevated height without a given angle can be determined by finding the vertex of the parabolic path that the object follows. This can be done by using the equation for the maximum height of a projectile, which is half the product of the initial vertical velocity and the time of flight.