A shell is shot with an inital velocity of 20m/s at an angle 60 degrees with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment whose speed is immediately after the explosion zero falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and the air drag is negligible?
Would I use v^2 =v_o^2 +2ad? and solve for d? but would I also use F=ma to find a? I really need some help. PLEASE help me out...
didn't I just do this?
http://www.jiskha.com/display.cgi?id=1170206253
15 meter
To solve this problem, you can use the kinematic equations and the principle of conservation of momentum.
First, let's break down the problem into different parts:
1. Initial motion of the shell: The shell is shot with an initial velocity of 20 m/s at an angle of 60 degrees with the horizontal. You can use the basic kinematic equations to find the initial vertical and horizontal components of the velocity.
- Vertical component: v_y = v_o * sin(theta)
- Horizontal component: v_x = v_o * cos(theta)
Here, v_o represents the initial velocity of 20 m/s and theta is the angle of 60 degrees.
2. Motion after explosion: At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment falls vertically with zero initial velocity.
Since the one fragment falls vertically with no initial velocity, we can conclude that the vertical component of the velocity of the other fragment immediately after the explosion is equal to the initial vertical component of the shell, which is v_y = v_o * sin(theta).
Now, let's apply the principle of conservation of momentum to find the horizontal component of the velocity of the other fragment after the explosion.
Conservation of Momentum:
m_shell * v_shell = m_fragment * v_fragment
Since the fragments have equal mass, we can simplify this equation to:
m_shell * v_shell = m_fragment * v_fragment
v_fragment = (m_shell / m_fragment) * v_shell
Here, m_shell represents the mass of the shell and m_fragment represents the mass of each fragment.
3. Final motion of the other fragment: Now, you can use the kinematic equations to find the distance traveled by the other fragment horizontally. Since the vertical component of its velocity is zero, the only force acting on it would be gravity in the downward direction. Neglecting air drag, the horizontal motion of the fragment is unaffected.
The horizontal distance traveled by the other fragment can be calculated using the equation:
d = v_x * time
Since the time of flight of the shell is the same as the time taken by the other fragment to reach the ground vertically, you can use the equation:
time = (2 * v_y) / g
Here, g represents the acceleration due to gravity.
Once you have calculated the time, you can substitute it in the earlier equation to find the horizontal distance traveled by the other fragment.
So, to summarize:
1. Find the initial vertical and horizontal components of the velocity of the shell.
2. Use the conservation of momentum to find the horizontal component of the velocity of the other fragment.
3. Use the kinematic equations to find the time of flight.
4. Substitute the time in the horizontal distance equation to find the distance traveled by the other fragment.
I hope this explanation helps! Let me know if you need any further assistance.