Angled Projectile at an elevated height

In summary, the conversation discusses the use of game physics in modern 3D video games and the task of shooting a cannon at a distant target. The target is at a distance of 400 m and an elevation of 150 m, with a launch speed of 100 m/s. The equations of motion are mentioned, specifically the incorrect equation for displacement. The conversation then moves onto finding the angles at which the cannon should be fired to hit the target, with the larger angle being 77.6 degrees and the smaller angle being 12.4 degrees.
  • #1
darkdeeds
7
0

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?



Homework Equations


1/cos^2(x)=tan^2(x)+1
v=d/t
Vf=Vo+at
d=Vot+(1/2)at^2
d=(Vf^2+Vo^2)/2a
a=9.81 m/s^2


The Attempt at a Solution



I don't even know where to start, I'm aware that there are 2 angle with the difference being 90-x(angle).
 
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  • #2
hi darkdeeds, welcome to physicsforums :) Is this part of homework? Anyway, I think your 5th equation is not quite right. It should be:
d=(Vf^2-Vo^2)/2a
And about the question itself, have you done these types of question before? What do you know about finding the components of velocity in vertical and horizontal directions? And what do you know about how the equations of motion work for vertical and horizontal motion?
 
  • #3
I've had similar questions to ones like this, however, I'm having difficulty finding the value for theta, and am not sure on how it is to be calculated without the time given. We've had questions like this before but they supplied more information.
 
  • #4
I was able to figure out the larger angle, which is 77.6 degrees, I am instructed to find the smaller angle, and I'm not having any luck, usually it should just be 90-77.6 to get 12.4 but the answer is wrong.
 
  • #5


I would approach this problem by first defining the variables and parameters involved. In this case, we have a cannonball being launched at a target located at a distance of 400 m and a height of 150 m. The launch speed of the cannonball is 100 m/s relative to the player's avatar, and we are assuming no wind resistance.

Next, I would use the equations for projectile motion to solve for the angles at which the cannon should be fired in order to hit the target. The first step would be to find the time of flight, which can be calculated using the equation d=Vot+(1/2)at^2. In this case, the initial velocity (Vo) is 100 m/s and the final distance (d) is 400 m. We can assume that the acceleration (a) is equal to the acceleration due to gravity (9.81 m/s^2).

Using this equation, we can solve for the time of flight, which comes out to be approximately 6.47 seconds.

Next, we can use the equation Vf=Vo+at to find the vertical component of the velocity at the time of impact. We know that the final velocity (Vf) will be equal to 0 at the time of impact, and the initial velocity (Vo) is 100 m/s. We can also assume that the acceleration (a) is equal to the acceleration due to gravity (9.81 m/s^2).

Using this equation, we can solve for the time at which the cannonball reaches its maximum height, which comes out to be approximately 3.23 seconds.

Now, we can use the equation d=(Vf^2+Vo^2)/2a to solve for the horizontal distance traveled by the cannonball in 3.23 seconds. This distance will be equal to the horizontal distance between the player and the target.

Once we have this distance, we can use the trigonometric equation 1/cos^2(x)=tan^2(x)+1 to solve for the angle at which the cannon should be fired in order to hit the target. We know that the horizontal distance is equal to 400 m, and the vertical distance is equal to 150 m. Using these values, we can solve for the angle at which the cannon should be fired.

In conclusion, as a scientist, I would use the equations for projectile motion and trigonometry to solve
 

Related to Angled Projectile at an elevated height

1. What is an angled projectile at an elevated height?

An angled projectile at an elevated height refers to an object that is launched at an angle from a certain height above the ground. This type of motion is influenced by both the initial velocity and the angle at which the object is launched.

2. How is the trajectory of an angled projectile at an elevated height affected by the initial velocity?

The initial velocity of an angled projectile at an elevated height determines the speed and direction of the object's motion. A higher initial velocity will result in a greater horizontal distance traveled by the object, while a lower initial velocity will result in a shorter horizontal distance.

3. What factors affect the maximum height reached by an angled projectile at an elevated height?

The maximum height reached by an angled projectile at an elevated height is affected by the initial velocity, the angle of launch, and the acceleration due to gravity. A higher initial velocity and a lower angle of launch will result in a greater maximum height, while a lower initial velocity and a higher angle of launch will result in a lower maximum height.

4. How does air resistance affect the motion of an angled projectile at an elevated height?

Air resistance, also known as drag, will decrease the distance traveled by an angled projectile at an elevated height. This is because air resistance acts in the opposite direction of the object's motion, slowing it down and reducing its maximum height and horizontal distance.

5. Can an angled projectile at an elevated height reach the same height as its launch point?

Yes, if the initial velocity and angle of launch are just right, an angled projectile at an elevated height can reach the same height as its launch point. This is known as the maximum height or the peak of the projectile's trajectory. However, this is only possible in an ideal scenario without any external factors such as air resistance.

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