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?
The fragment whose speed is zero, must have had momenum applied to it in the horizontal direction to "stop". That momenum is - 1/2 m vhorizontal, as it was going vhorizonal at the time. That means the other piece has 1/2 m vhorizontal applied to it, and it moves with m vhorizontal forward.
So, since it's mass is 1/2 m, it now has twice the horizontal speed (2*20cos60).
Figure the time it takes to get to the top. Then it takes the same time tofall, but the bulliet piecce is going twice as fast.
To solve this problem, let's break it down step by step:
Step 1: Calculate the time it takes for the shell to reach its maximum height.
To do this, we can use the equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the final velocity at the top of the trajectory is zero, the initial velocity is 20 m/s, and the acceleration is due to gravity, which is approximately -9.8 m/s^2 (taking downward as negative).
0 = 20sin(60) - 9.8t
Simplifying the equation, we get:
9.8t = 20sin(60)
t = 20sin(60)/9.8
Step 2: Calculate the horizontal distance traveled by the shell before it explodes.
We need to find the horizontal component of the initial velocity, which can be calculated using the equation: v_horizontal = u_horizontal + a_horizontal*t.
Given that the initial velocity is 20 m/s and the angle with the horizontal is 60 degrees, we can calculate the horizontal component of the initial velocity:
u_horizontal = u * cos(60) = 20 * cos(60)
Step 3: Calculate the horizontal distance traveled by the other fragment.
Since the fragment with zero speed falls vertically, it does not contribute to the horizontal distance. The other fragment, with twice the horizontal speed (2 * u_horizontal), covers the entire horizontal distance.
Step 4: Calculate the total horizontal distance traveled by the other fragment.
The total horizontal distance traveled by the other fragment can be calculated using the equation: d = u_horizontal * t + (1/2) * a_horizontal * t^2. However, since the acceleration in the horizontal direction is zero (assuming no air drag), the equation simplifies to:
d = u_horizontal * t
Plugging in the values for u_horizontal and t, we get:
d = (20 * cos(60)) * (20 * sin(60) / 9.8)
Simplifying further, we have:
d = (20 * cos(60) * 20 * sin(60)) / 9.8
Finally, calculate the value of d to find the horizontal distance traveled by the other fragment.