# Suppose a 10.0 kg fireworks shell is shot into the air with an initial velocity of 62.0 m/s at an angle of 80.0 degrees above the horizontal. At the highest point of its trajectory, a small explosive charge separates it into two pieces, neither of which ignite (two duds). One 9.00 kg piece falls straight down, having zero velocity just after the explosion.

(a) At what horizontal distance from the starting point does the 9.00 kg piece hit the ground?
(b) Calculate the velocity of the 1.00 kg piece just after the separation.
(c) At what horizontal distance from the starting point does the 1.00 kg piece hit the ground?

**** I apologize in advance for not having any work to show, but its because I have no clue on how to approach this problem. If I could get some help on what equations to use, that would help.

## atart by fining the initial vertical velocity, and the initial horizontal velocity.

Then, figure the highest point, at which velocity is zero.
Then, note the velocity of the duds. One is zero. At the highest point, the only momentum is due to horizonal velocity, so dud number two must be going now twice the original horizonal velocity.
So find:
a) highest point, and time of hightest point.
b) horizontal distance from launch to hightest point.
c) now find the time for the dud to fall from the hightest point, and then the horizonal distance during falling is d=2vihoriz*timefalling

now add the two horizonal distances, and you can take a break

To solve this problem, we can break it down into several steps:
1. Analyze the motion of the fireworks shell before the explosion.
2. Determine the vertical and horizontal components of the velocity of the 9.00 kg piece after the explosion.
3. Use the vertical motion to calculate the time it takes for the 9.00 kg piece to hit the ground.
4. Use the horizontal motion to calculate the horizontal distance traveled by the 9.00 kg piece.
5. Calculate the velocity of the 1.00 kg piece just after the separation.
6. Use the vertical motion to calculate the time it takes for the 1.00 kg piece to hit the ground.
7. Use the horizontal motion to calculate the horizontal distance traveled by the 1.00 kg piece.

Now let's go step by step.

Step 1: Analyze the motion of the fireworks shell before the explosion.
The initial velocity of the fireworks shell is given as 62.0 m/s at an angle of 80.0 degrees above the horizontal. To analyze its motion, we need to break this initial velocity into its vertical and horizontal components.

The vertical component is given by V_y = V * sin(θ), where V is the magnitude of the initial velocity and θ is the angle of the initial velocity above the horizontal.
So, V_y = 62.0 m/s * sin(80.0 degrees).

The horizontal component is given by V_x = V * cos(θ), where V is the magnitude of the initial velocity and θ is the angle of the initial velocity above the horizontal.
So, V_x = 62.0 m/s * cos(80.0 degrees).

Step 2: Determine the vertical and horizontal components of the velocity of the 9.00 kg piece after the explosion.
Before the explosion, the 10.0 kg fireworks shell had a single velocity vector. However, after the explosion, the 9.00 kg piece will have a different velocity.

The vertical component of the velocity of the 9.00 kg piece just after the explosion is the same as the vertical component of the velocity of the entire shell just before the explosion, since the explosion does not change this component.

For the horizontal component, it is important to note that no horizontal forces act on the 9.00 kg piece after the explosion. This means that there is no change in the horizontal velocity component. Therefore, the horizontal component of the velocity of the 9.00 kg piece will remain the same as that of the entire shell just before the explosion.

Step 3: Use the vertical motion to calculate the time it takes for the 9.00 kg piece to hit the ground.
The motion of the 9.00 kg piece after the explosion can be analyzed as vertical projectile motion. We can use the equation for the vertical displacement of a projectile:

Δy = V_y * t + (1/2) * g * t^2,

where Δy is the vertical displacement, V_y is the vertical component of velocity, t is the time, and g is the acceleration due to gravity.

Since the 9.00 kg piece falls straight down, its initial vertical velocity is zero. Therefore, the equation simplifies to:

Δy = (1/2) * g * t^2.

We know that the vertical displacement Δy is the height from which the 9.00 kg piece falls straight down.

Step 4: Use the horizontal motion to calculate the horizontal distance traveled by the 9.00 kg piece.
Since there are no horizontal forces acting on the 9.00 kg piece after the explosion, it will continue to move with the same horizontal velocity as that of the fireworks shell just before the explosion. Therefore, we can use the equation for horizontal motion:

Δx = V_x * t,

where Δx is the horizontal distance traveled and V_x is the horizontal component of velocity.

Step 5: Calculate the velocity of the 1.00 kg piece just after the separation.
The velocity of the 1.00 kg piece just after the separation can be calculated using the principle of conservation of momentum. Since there are no external horizontal forces acting on the system (the fireworks shell and its pieces), the horizontal component of momentum is conserved. Therefore, we have:

(m1 * V1)_x + (m2 * V2)_x = (m1 * V1')_x + (m2 * V2')_x,

where m1 and m2 are the masses of the two pieces, V1 and V2 are the horizontal components of the initial velocities of the shell and the 9.00 kg piece, V1' and V2' are the horizontal components of the final velocities of the 1.00 kg piece and the 9.00 kg piece, respectively.

Since the 9.00 kg piece falls straight down after the explosion, its horizontal velocity component V2' will be zero. The initial velocity V1 becomes the final velocity V1' of the 1.00 kg piece. Therefore, we can solve the equation for V1'.

Step 6: Use the vertical motion to calculate the time it takes for the 1.00 kg piece to hit the ground.
Similar to the 9.00 kg piece, the motion of the 1.00 kg piece after the separation can be analyzed as vertical projectile motion. We can use the equation:

Δy = V_y * t + (1/2) * g * t^2,

where Δy is the vertical displacement, V_y is the vertical component of velocity, t is the time, and g is the acceleration due to gravity.

Since the 1.00 kg piece falls straight down, its initial vertical velocity is zero. Therefore, the equation simplifies to:

Δy = (1/2) * g * t^2.

We know that the vertical displacement Δy is the height from which the 1.00 kg piece falls straight down.

Step 7: Use the horizontal motion to calculate the horizontal distance traveled by the 1.00 kg piece.
Since there are no horizontal forces acting on the 1.00 kg piece after the separation, it will continue to move with the same horizontal velocity as that of the 9.00 kg piece just before the separation. Therefore, we can use the equation for horizontal motion:

Δx = V_x * t,

where Δx is the horizontal distance traveled and V_x is the horizontal component of velocity.

By following these steps, you should be able to find the answers to parts (a), (b), and (c) of the problem.