- #1

Riman643

- 62

- 2

- Homework Statement
- A shell is shot with an initial velocity Vo of 22 m/s, at an angle of θ = 59° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass (see the figure). One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that air drag is negligible?

- Relevant Equations
- Basic kinematics

p = mv

Okay so the first thing I did was was find the height of the apex. I found this by finding the x and y components of Vo:

Vox = VoCos(θ) = 22cos(59) = 11.33 m/s

Voy = VoSin(θ) = 22sin(59) = 18.86 m/s

Using the initial velocity of the y component I found the time it took to reach the apex:

t = -Voy / ay = -18.86/-9.8 = 1.93 s

Using the time I found the height of the apex:

Δy = Voyt + (1/2)ayt^2 = 18.86(1.93) + (1/2)(-9.8)(1.93^2) = 18.15 m

Using the height I found the time it took after the explosion for the bullet fragment to reach the ground. Because the bullet was at the apex, the initial velocity of the fragments is zero in the y direction

t = √(Δy/(1/2ay)) = √[(-18.15)/((1/2)(-9.8))] = 1.94s

Using the momentum equation I was able to find the initial velocity of the bullet fragment in the x direction:

Pi = Pf; 22m = (1/2)mVo; Vo = 44

Finally using the two times and two speeds of the fragment in the x direction I found (wrong according to the website) the total distance:

Δx = (Vox * t) + (Vox * t) = (11.33)(1.93) + (44)(1.94) = 107.23 m

Vox = VoCos(θ) = 22cos(59) = 11.33 m/s

Voy = VoSin(θ) = 22sin(59) = 18.86 m/s

Using the initial velocity of the y component I found the time it took to reach the apex:

t = -Voy / ay = -18.86/-9.8 = 1.93 s

Using the time I found the height of the apex:

Δy = Voyt + (1/2)ayt^2 = 18.86(1.93) + (1/2)(-9.8)(1.93^2) = 18.15 m

Using the height I found the time it took after the explosion for the bullet fragment to reach the ground. Because the bullet was at the apex, the initial velocity of the fragments is zero in the y direction

t = √(Δy/(1/2ay)) = √[(-18.15)/((1/2)(-9.8))] = 1.94s

Using the momentum equation I was able to find the initial velocity of the bullet fragment in the x direction:

Pi = Pf; 22m = (1/2)mVo; Vo = 44

Finally using the two times and two speeds of the fragment in the x direction I found (wrong according to the website) the total distance:

Δx = (Vox * t) + (Vox * t) = (11.33)(1.93) + (44)(1.94) = 107.23 m