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?

To find the horizontal distance traveled by the other fragment, we can follow these steps:

1. First, we need to find the time it takes for the shell to reach its maximum height. We can use the vertical motion equation for projectile motion:
h = (v₀*sinθ)² / (2*g)
where h is the maximum height, v₀ is the initial velocity (20 m/s), θ is the launch angle (60 degrees), and g is the acceleration due to gravity (9.8 m/s²). Plugging in these values, we get:
h = (20*sin60°)² / (2*9.8) ≈ 10.125 m

2. The time taken to reach the maximum height can be found using the equation:
t = v₀*sinθ / g
Plugging in the values, we get:
t = 20*sin60° / 9.8 ≈ 1.15 s

3. Since the shell explodes at the top of its trajectory, the vertical component of velocity immediately after the explosion is zero. We can use this to find the time taken for the shell to reach the top and start falling back down:
t_total = 2*t ≈ 2.3 s

4. Now, we need to find the horizontal component of velocity of the shell. We can use the equation:
v_x = v₀*cosθ
Plugging in the values, we get:
v_x = 20*cos60° ≈ 10 m/s

5. Finally, we can find the horizontal distance traveled by the other fragment using:
d = v_x * t_total
Plugging in the values, we get:
d = 10 * 2.3 ≈ 23 m

Therefore, the other fragment would land approximately 23 meters away from the gun, assuming no air drag and level terrain.

First break up V initial into components. Then you must calculate the time it takes for the shell to reach it's peak using Vy=V0y-gt (should get 1.77s). Then, find displacement in the x before the explosion happens using x=Vox*t (since there's no acceleration in the x-direction, the product to the right of the sum goes to zero). We get 17.7m in the x. Then we can use the formula for center of mass to find where the other fragment lands.