Need help Projectile motion after breaking in two pieces

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SUMMARY

The discussion centers on a projectile motion problem where an object launched from Earth breaks into two pieces at a horizontal distance of 300 meters. The smaller piece, with 1/4 of the mass, falls straight down, while the larger piece, with 3/4 of the mass, continues to travel. The conclusion drawn is that the larger piece lands at a horizontal distance of 600 meters from the launch point, as it maintains the trajectory of the center of mass of the system.

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  • Familiarity with 1-D kinematics
  • Basic algebra for solving equations
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Koskesh
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Homework Statement


An object is launched from Earth as a projectile. At the top of its trajectory the object breaks into separate pieces at a horizontal distance 300m from the launch point. One piece has 1/4 of the mass of the initial object and the other piece has 3/4 of the mass of the initial object. The smaller piece falls straight to the ground while the larger continues its motion.

At what horizontal distance from the launch point does the larger piece land?

Homework Equations


Do I need to use a projectile motion equation?
M1 = 1/4m
M2 = 3/4m
Vy = 0 m/s

The Attempt at a Solution


I know that both pieces will hit the ground at the same time b/c of 1-D kinematics.
I'm so lost but here's an attempt:
For 1/4m = M1
(M1)gh = 1/2mv^2
(1/4m)gh = 1/2(1/4m)v^2
Okay now I don't have height or anything??
 
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Koskesh said:

Homework Statement


An object is launched from Earth as a projectile. At the top of its trajectory the object breaks into separate pieces at a horizontal distance 300m from the launch point. One piece has 1/4 of the mass of the initial object and the other piece has 3/4 of the mass of the initial object. The smaller piece falls straight to the ground while the larger continues its motion.

At what horizontal distance from the launch point does the larger piece land?

Homework Equations


Do I need to use a projectile motion equation?
M1 = 1/4m
M2 = 3/4m
Vy = 0 m/s

The Attempt at a Solution


I know that both pieces will hit the ground at the same time b/c of 1-D kinematics.
I'm so lost but here's an attempt:
For 1/4m = M1
(M1)gh = 1/2mv^2
(1/4m)gh = 1/2(1/4m)v^2
Okay now I don't have height or anything??

Maybe you are making it too complicated?

If it didn't break apart the whole thing would land another 300m further on wouldn't it? So when it breaks apart where does the center of mass end up? If you guessed 300m I think that would be a good guess. So ... if the center of mass is at 600m and the 1/4 piece is at 300m mark where would you look for the 3/4 piece?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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