How Far Does the Second Fragment Travel from Point A After Explosion?

In summary: It turns out to be this. V1/e= 15m/s and V2/e= 5m/s. They would have different velocities relative to the earth. Doesn't that change the time for them to come down to the ground?The x component of the velocities are different, but the y components are the same. The parts descend together, but cover different horizontal distances.
  • #1
Toranc3
189
0

Homework Statement



A projectile is fired from point A at an angle above the horizontal. At its highest point, after having travled a horizontal distance D from its launch point, it suddenly explodes into 2 identical fragments that travel horizonatally with equal but opposite velocitites as measured relative to the projectile just before it exploded. If one fragment lands back at point A, how far from A (in terms of D) does the other fragment land?

Homework Equations



V1/e-x=V1/p-x + Vp/e-x


The Attempt at a Solution


1 is going to the right(+)
V1/e=V1/p + Vp/e

V1/e = velocity of 1 relative to earth.
V1/p= velocvity of 1 relative to projectile original
Vp/e= velocity of projectile relative to earth

2 is going to the left(-)
V2/e-x=V2/p-x + Vp/e-x

V2/e = velocity of 2 relative to earth.
V2/p= velocvity of 2 relative to projectile original
Vp/e= velocity of projectile relative to earth


This is what I have for one

X=Xo +vox*t

X=D+(V1/p + Vp/e)*t

Ok so at first I thought that the time it took for the 2nd to fall down to origin(A) was the same time it took the first to come to the ground. Then I saw it different. I realized that the 1st fragment(+) actually has a higher velocity after explosion relative to the earth. So that must mean that it could have taken longer for it to fall to the ground. I am stuck here though and I was wondering if I could get some hints.
 
Last edited:
Physics news on Phys.org
  • #2
What are the y components of velocity of the parts, after the explosion? Was the projectile fired from the ground?ehild
 
  • #3
ehild said:
What are the y components of velocity of the parts, after the explosion? Was the projectile fired from the ground?


ehild

The y components are zero. I am assuming that the projectile was fired from the ground.
 
  • #4
ehild said:
What are the y components of velocity of the parts, after the explosion? Was the projectile fired from the ground?


ehild

Still a bit confused.
 
  • #5
Both parts starts to fall from the same height, with zero vertical velocity. Any reason why they do not reach the ground at the same time?

ehild
 
  • #6
ehild said:
Both parts starts to fall from the same height, with zero vertical velocity. Any reason why they do not reach the ground at the same time?

ehild

Yeah that makes sense. That is what I originally thought. I would think that the answer would be 2D.

But then this came up.

V1/e-x = V1/p-x+ Vp/e-x and this V2/e-x=V2/p-x + Vp/e-x

V1/p and V2/p are both equal in magnitude but opposite in direction right?
Let us say that it is 5 and vp/e is 10.

It turns out to be this. V1/e= 15m/s and V2/e= 5m/s. They would have different velocities relative to the earth. Doesn't that change the time for them to come down to the ground?
 
Last edited:
  • #7
The x component of the velocities are different, but the y components are the same. The parts descend together, but cover different horizontal distances.

ehild
 

Related to How Far Does the Second Fragment Travel from Point A After Explosion?

1. What is a projectile fired from point A?

A projectile fired from point A is an object that is launched into the air from a specific starting point, known as point A. The object follows a curved path, known as a trajectory, under the influence of gravity until it reaches its highest point and then falls back down to the ground.

2. What factors affect the trajectory of a projectile fired from point A?

The trajectory of a projectile fired from point A is affected by several factors, including the initial velocity of the object, the angle at which it is launched, air resistance, and the force of gravity. The shape and weight of the object can also play a role in its trajectory.

3. How is the distance traveled by a projectile fired from point A calculated?

The distance traveled by a projectile fired from point A is calculated using the equation: d = v0 * t * cos(theta). Where d is the distance, v0 is the initial velocity, t is the time in flight, and theta is the angle at which the object is launched.

4. What is the maximum height reached by a projectile fired from point A?

The maximum height reached by a projectile fired from point A is determined by the initial velocity and angle of launch. It can be calculated using the equation: h = (v0*sin(theta))^2 / 2g, where v0 is the initial velocity, theta is the angle of launch, and g is the acceleration due to gravity (9.8 m/s^2).

5. How does air resistance affect a projectile fired from point A?

Air resistance, also known as drag, can affect the trajectory of a projectile fired from point A by slowing it down and changing its direction. This can cause the object to fall short of its expected distance or deviate from its intended path. Objects with a larger surface area, such as a feather, will be more affected by air resistance than objects with a smaller surface area, such as a bullet.

Similar threads

Replies
13
Views
1K
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
18
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
7K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
22
Views
3K
Back
Top