- #1

kashmirekat

- 30

- 1

Two masses are each initially 1.80 m above the ground, and the massless frictionless pulley is 4.8m above the ground. What maximum height does the lighter object reach after the system is released?

It gave me a hint to find the acceleration and velocity at the moment the heavier one hit the ground.

To find 'a' I used ((M1-M2)/(M1+M2))*9.8m/s^2 and got the answer 1.815m/s^2.

For velocity, I used the equation V=Vo+at, where time was determined by the distance to the ground, 1.8m divided by acceleration 1.815m/s^2. t=1.8/1.815m/s^2 and I got t=0.99s. Question here, since the a is squared, I would have to square my t, right?

t=sqrt(d/a).

Now my real question is what does it want for maximum height? The distance to the pulley is 4.8m...and the heavier object is only going to travel 1.8m downward until it reaches the ground. So thinking like a lamen, maximum height upward should only be 3m....but it's more complicated than that, isn't it? Why and where do I need to use my determined acceleration and velocity?