Finding the Location of a Projectile After 3 Seconds

In summary: We're done.If you want to find the angle, you need to know that tan θ = Δy/Δx. You know both of those, so you can find θ.In summary, an object is shot from the origin with a velocity of 40.0 m/s at an angle of 55.0 degrees above the horizontal. After 3.00 seconds, the object will have traveled a horizontal distance of 68.8 meters and a vertical distance of -55.4 meters. The location of the object after 3.00 seconds is (68.8, -55.4). To find the angle of the object's velocity vector, you can use the tangent function with the horizontal
  • #1
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Homework Statement



An object is shot from the origin with a velocity of 40.0 m/s at an angle of 55.0 degrees above the horizontal. What is the location of the object 3.00 seconds later?


Homework Equations



All 2D motion equations - too many to list Ex:

[tex] V_x = v_0x + a_x t [/tex]

[tex] V_y = v_0y + a_y t [/tex]


The Attempt at a Solution


What I am having trouble with is the 3.0 seconds later part. I am not sure how to approach this one. I have to find the x displacement and then the angle [tex] \theta [/tex] which is no problem to do when I know [tex] V_y and V_x [/tex].

So I tried this:

[tex] \Delta x = 40 cos(55)3.0 = 68.8m [/tex] , but none of the answer choices match this. So I must be doing something wrong. I also can't get a correct answer for [tex] V_y and V_x [/tex] with these I could plug them into the inverse tangent equation and get an angle. To find [tex] V_x [/tex] I do this:

[tex] V_x = 40 cos (55) = 22.9 [/tex]

for [tex] V_y [/tex] I do this :

[tex] V_y = 40 sin (55) - (9.80)(3.0^2) = -55.4 [/tex]

I am not sure that I am using the time variable correctly.
 
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  • #2
Want to learn said:

Homework Statement



An object is shot from the origin with a velocity of 40.0 m/s at an angle of 55.0 degrees above the horizontal. What is the location of the object 3.00 seconds later?

Homework Equations



All 2D motion equations - too many to list Ex:

[tex] V_x = v_0x + a_x t [/tex]

[tex] V_y = v_0y + a_y t [/tex]

The Attempt at a Solution


What I am having trouble with is the 3.0 seconds later part. I am not sure how to approach this one. I have to find the x displacement and then the angle [tex] \theta [/tex] which is no problem to do when I know [tex] V_y and V_x [/tex].

So I tried this:

[tex] \Delta x = 40 cos(55)3.0 = 68.8m [/tex] , but none of the answer choices match this. So I must be doing something wrong. I also can't get a correct answer for [tex] V_y and V_x [/tex] with these I could plug them into the inverse tangent equation and get an angle. To find [tex] V_x [/tex] I do this:

[tex] V_x = 40 cos (55) = 22.9 [/tex]

for [tex] V_y [/tex] I do this :

[tex] V_y = 40 sin (55) - (9.80)(3.0^2) = -55.4 [/tex]

I am not sure that I am using the time variable correctly.

You can solve the horizontal and vertical parts independently.

Your Δx of 68.8 looks fine.

Your final equation for Vy is wrong. Are you calculating a velocity, or a distance?

Under relevant equations, you've given a correct formula for getting a velocity, given initial velocity and acceleration... but it doesn't have a t2 factor anywhere. The question, of course, requires a distance.

Do you know an equation for distance, given an initial velocity, an acceleration, and a time?

Cheers -- sylas
 
  • #3
well there is this one:

[tex] \Delta x = V_0_x t + (1/2) (a_x) t^2 [/tex]

but [tex] a [/tex] in the horizontal direction = 0 in this case. So I am back to where I started.

I am trying to find, where the object is going to end up after 3.00 seconds.

I need the distance traveled in 3 seconds. And then I need to find the angle that vector makes with the x-axis using the inverse tangent equation. But to use the tangent equation I need to find [tex] V_y and V_x [/tex]

These are the equations from my book:

[tex] V_y = V_0 sin \theta - gt [/tex]
[tex] V_x = V_0 cos \theta [/tex]

when I use these, it doesn't work out.
 
  • #4
Want to learn said:
well there is this one:

[tex] \Delta x = V_0_x t + (1/2) (a_x) t^2 [/tex]
That correctly gave you x=68.8 m. If you use the analogous equation for Δy, you'll get y as well.

Since the question asks for the location of the object, all you need is x and y. We don't need Vx, Vy, or the distance travelled.
 
2D Motion part deux FAQ

FAQ: 2D Motion part deux

What is 2D Motion part deux?

2D Motion part deux refers to the motion of an object in a two-dimensional space, taking into account both its horizontal and vertical components.

What are the equations for 2D Motion part deux?

The equations for 2D Motion part deux include the equations for displacement, velocity, and acceleration in both the horizontal and vertical directions. These equations are similar to those used in 1D motion, but with separate components for each direction.

What is the difference between 2D Motion part deux and regular 2D motion?

In regular 2D motion, the object moves in a two-dimensional space without taking into account any other forces acting on it. In 2D Motion part deux, external forces such as friction and air resistance are also considered in addition to the motion of the object.

How is 2D Motion part deux used in real life?

2D Motion part deux is used in various fields such as engineering, physics, and video game development to accurately model the motion of objects in a two-dimensional space. It is also used in sports analysis to track the movement of athletes.

What are some common misconceptions about 2D Motion part deux?

One common misconception is that the object's motion in the horizontal and vertical directions are completely independent of each other. In reality, the object's motion in one direction can affect its motion in the other direction due to external forces and acceleration.

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