A shell is fired from a gun with a muzzle velocity of 466m/s at an angle of 57.4 degrees with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragments lan assumin level terrain?
To find the horizontal distance traveled by the other fragment, we first need to determine the time it takes for the shell to reach the top of its trajectory.
1. Split the initial velocity into its vertical and horizontal components:
- Vertical component: V₀y = V₀ * sin(θ)
- Horizontal component: V₀x = V₀ * cos(θ)
2. Calculate the time it takes for the shell to reach the top of its trajectory using the vertical component of velocity:
- Time to reach the top: t = V₀y / g
(where g is the acceleration due to gravity, approximately 9.8 m/s²)
Now that we know the time it takes for the shell to reach the top, we can find the total time of flight.
3. The total time of flight (T) for a projectile fired at an angle can be found using the equation:
- T = 2 * t
Once we have the total time of flight, we can determine the horizontal distance traveled by the other fragment.
4. Calculate the horizontal distance (d) traveled by the shell using the horizontal component of velocity and the total time of flight:
- d = V₀x * T
Now we can plug in the values to find the answer:
Given:
- Muzzle velocity (V₀) = 466 m/s
- Launch angle (θ) = 57.4 degrees
- Mass of shell fragment = Mass of falling fragment
V₀y = V₀ * sin(θ)
V₀x = V₀ * cos(θ)
t = V₀y / g
T = 2 * t
d = V₀x * T
Calculate:
V₀y = 466 m/s * sin(57.4°)
V₀x = 466 m/s * cos(57.4°)
t = V₀y / 9.8 m/s²
T = 2 * t
d = V₀x * T
Finally, calculate the values to find the answer.