Projectile Problem: Solving an Artillery Shell Problem

[Zdub311]

Homework Statement

An artillery shell is fired from the origin in the x direction(z is vertically upward) at an angle of 30degrees above the horizontal axis at a velocity of 3000 ft/s. after 2.5a the shell which weighs 100lb explodes into two pieces. The first piece to land weighs 30 lbs and lands at (900,400,0)ft at 4.5 s after the gun was fired. When and where does the other piece land? Neglect air resistance and assume

Homework Equations

Not sure on these but I have x_e=v cos theta(t) and z_e=v sin theta(t)_e -1/2gt_e^2 and t_e=o

The Attempt at a Solution

Really need some inspiration with this one guys. This is my first physics class btw!

Thank you for any attempted help.

 

[Zdub311]

Although a 3000 level undergrad physics course. This question probably belongs in the beginner forum!

 

[nejibanana]

I would use the kinematic equations, I think they are what you have listed, the notation throws me off a bit. Using those you should be able to find the velocity of the particle at the point it explodes, and also find out the velocity and angle of the first piece. From there conservation of momentum should be able to give you the velocity and angle for the second piece, and using the kinematic equations once again will give you what it asks for. This assumes the explosion doesn't impart any energy into the pieces.

 

[AlephZero]

This assumes the explosion doesn't impart any energy into the pieces.

A change in kinetic energy doesn't matter. The assumption is that the two pieces contain all the mass of the shell. That means you can use conservation of momentum at the time when the shell explodes.

Find where the shell is at 2.5s.
Then find the velocity vector after the explosion for the 30lb piece, so it lands at the given time and place.
Then use conservation of momentum to find the velocity vector of the other piece after the explosion.

 

[HalfLostAndSearching]

I would approach this problem by first identifying and organizing the given information. We know that an artillery shell weighing 100 lbs is fired at an angle of 30 degrees above the horizontal axis with a velocity of 3000 ft/s. After 2.5 seconds, it explodes into two pieces, with one piece weighing 30 lbs and landing at (900,400,0)ft at 4.5 seconds after the gun was fired. We are asked to determine when and where the other piece lands.

To solve this problem, we can use the equations of motion to calculate the position and time of the second piece. Since we are neglecting air resistance, we can use the equations for constant acceleration:

x = x0 + v0t + 1/2at^2
y = y0 + v0t + 1/2at^2
z = z0 + v0t + 1/2at^2

Where:
x, y, and z are the final positions in the x, y, and z directions respectively
x0, y0, and z0 are the initial positions in the x, y, and z directions respectively
v0 is the initial velocity
a is the acceleration
t is the time

Since the first piece lands at (900,400,0)ft at 4.5 seconds, we can use this information to determine the initial positions and time for the second piece. We can also use the fact that the total time of flight for the shell is 2.5 seconds, so the second piece will land at 7 seconds (2.5 seconds after the explosion).

Using the given initial velocity of 3000 ft/s and the angle of 30 degrees, we can find the initial velocities in the x and z directions using trigonometric functions. We can also use the given weight of the second piece (70 lbs) to calculate the acceleration due to gravity.

Once we have all the necessary information, we can plug it into the equations of motion to solve for the final position of the second piece:

x = 900 + v0x(7-4.5)
y = 400 + v0y(7-4.5)
z = 0 + v0z(7-4.5) - 1/2gt^2

Solving these equations will give us the final position of the second piece, which