Physics--pulley problem--distance traveled?

Question

m1 is 1 kg and m2 is 3.5 kg. They are suspended from a frictionless pulley 14.9 m above the ground. The two masses are held even at 1.75 m above the ground. When the masses are released, what is the maximum height that the lighter mass will reach? Help on this would be great. I figured out the acceleration by adding the sum of the forces on m1 to the sum of the forces on m2. Now what do I do to figure out the distance traveled by m1? Do I figure out the time it takes m2 to hit the ground and then do something with that time?

Answer 1

A force F will act upon both masses

F=W2-W1=(m1 + m2)g
The acceleration of m1 going down an m2 going up is a and it is

a=F/(m1+m2)

as soon as the m2 hits the ground a=0
The m1 continues to travel for time t2 until it stops V=0
V=0=V0 - gt2
and V0=? (this is the speed when a=0)
V0=at1 and since
h1=1.75=0.5 a(t1)^2
t1= sqrt(2h1/a)
so V0=a sqrt(2h1/a)
V0=sqrt(2 h1 a)

max height = h1+h2
h2=0.5 g (t2)^2
t2=V0/g

Now just compute...

Answer 2

i will not see the diagram you're making use of, yet you're given a time, preliminary speed, and additionally you got here upon the acceleration. d = vo * t + 0.5*a*t^2 : d = 0.5*a*t^2 d = 0.5*a million.633m/s^2 * (0.875s)^2 d = 0.714 m