Could someone explain to me how to figure out part C) Im pretty

>>326479
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v = u + a t
⇒ ( v - u ) / a = t ... [ transposing for t ];

but v = 0, when 50kg stops due to force of gravity (assuming it doesn't hit the pulley first)

⇒ u / a = t ..... [ NB: u and a are both negative ];

⇒ divide initial velocity by acceleration due to gravity ( "deceleration" if you prefer ) in order to find the time interval ( t ) during which the 50kg mass continues to rise, after the 150kg mass reaches the ground;

Now:

s = u t + 0.5 a t 2

where:
s is the distance you are trying to find;
u is the initial velocity as deduced in part (b);
t is the time deduced, (as described above);
a is the acceleration due to gravity;

..... and don't forget to use a negative value for that "acceleration", because it is actually slowing the mass down.

>>326508
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final rectilinear motion function should read:

s = u t + 0.5 a t^2

..... 4chan obviously doesn't like the superscript 2 that I used to signify t squared.

>>326508
To find time in the first bit are you using the acceleration of the first part then or the acceleration from gravity (9.81)?

>>326517
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Quote: "... are you using the acceleration of the first part then or the acceleration from gravity (9.81)? ";

use the acceleration due to gravity, because the 150kg mass has just impacted the ground, so its velocity is zero, and its acceleration is also zero (it does not bounce).

the only force now acting upon the 50kg is now the gravitational force that is slowing it down.

Please note that I have not checked your working in parts (a) and (b), as you are:
"... pretty sure ... (you've) ... done part A) and B) correct ...",

>>326522
Thank you i get it now, very helpful, so basically you find the time it continues to travel for using the final velocity from part B) but use it as the initial velocity this time (and the final velocity as 0) and then put them in the equation along with gravity (as a negative) to find the distance.

So inserting the velocity 5.86 you would get -5.85/-9.81 giving 0.597... seconds

then: s = 5.86 * 0.597... + 1/2 * -9.81 * 0.597... * 2 = -2.3595....? how come its a negative?

>>326522
>>326542
just clicked on that its t squared not all *2 so i got 1.7498.... that sounds more like it

>>326543
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..... I agree, BUT I recommend you call it:

s ≅ 1.75 m;

..... rather than 1.7498 which is just TOO "precise" to conform to reality unless you really believe your calculations are accurate to ± 50 μm !

..... and NEVER forget to include the units of measurement:

..... the units are every bit as important as the numerical value, assuming that is expressed to a credible precision.

Don't expect anybody else to assume correctly what units you intended them to infer from your purely numerical working.
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