Need help Projectile motion after breaking in two pieces

[Koskesh]

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??

 

[LowlyPion]

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?

 

[nlightner]

Yes, you will need to use a projectile motion equation to solve this problem. The equation you will need to use is the range equation, which calculates the horizontal distance traveled by a projectile. The equation is:

R = (v0^2/g) * sin(2θ)

Where R is the range, v0 is the initial velocity, g is the acceleration due to gravity, and θ is the launch angle.

In this problem, the initial velocity of the larger piece will be the same as the initial velocity of the whole object before it broke into two pieces. The launch angle will also be the same.

So, you can calculate the range for the larger piece using the equation above. The smaller piece will fall straight to the ground, so its horizontal distance traveled will be 0.

To find the total horizontal distance traveled by the larger piece, you can simply add the horizontal distance at which it broke into two pieces (300m) to the range calculated using the equation above. This will give you the total horizontal distance traveled by the larger piece before it lands.

Hope this helps!