Simple problem, but confused with acceleration of gravity

[mizzy]

Homework Statement

A silo is under construction and a large brick is being hoisted up at a constant speed of 3.2m/s. The rope breaks when the brick is at a height of 28m above the ground.

a) find the greatest height reached by the brick.

Homework Equations

i know which equation to use:

v^2 = vo^2 = 2a(x2-x1)

The Attempt at a Solution

At the greatest height, i know final velocity will be zero.

in the solution, a is negative. this is where i get confused...i thought a is negative therefore slowing down, but is a negative because it's going down?

 

[Doc Al]

v^2 = vo^2 = 2a(x2-x1)

That should be a plus sign:
v^2 = vo^2 + 2a(x2-x1)

in the solution, a is negative. this is where i get confused...i thought a is negative therefore slowing down, but is a negative because it's going down?

Gravity acts down. If you use the standard convention of up = positive, then a will be negative. If something is moving upward, it will slow down; if something is moving downward, it will speed up. (When the signs--which indicate the direction--of v and a are the same, the object is speeding up; when the signs are opposite, it's slowing down.)

 

[mizzy]

That should be a plus sign:
v^2 = vo^2 + 2a(x2-x1)


Gravity acts down. If you use the standard convention of up = positive, then a will be negative. If something is moving upward, it will slow down; if something is moving downward, it will speed up. (When the signs--which indicate the direction--of v and a are the same, the object is speeding up; when the signs are opposite, it's slowing down.)

ok, i tried using that convention, but i get the wrong answer. Going up i used a positive 'a' and if it's falling down, i use a negative 'a'

For example for this question, i got it wrong =(

A projectile is launched straight up at 60.0m/s from a height of 80m at the edge of a cliff. The projectile falls, just missing the cliff and hitting the ground below. Find a) max height of the projectile above the point of firing.

In order to find the max height, i calculated the time it took to get to the max height, where v = 0. this is what i did:

v = vo +at
0 = 60 + 9.8t
t = -9.18s

then i solved for the max height using the equation: y = voyt + 1/2at^2

is that correct??

 

[iamalexalright]

you have a negative time there !

Remember that the way we have it set up the acceleration is negative

v = v_{0} + at
0 = 60 -gt
-60 = -gt
60 = gt
t = 60/g

 

[Doc Al]

ok, i tried using that convention, but i get the wrong answer. Going up i used a positive 'a' and if it's falling down, i use a negative 'a'

No, that's not the convention. 'a' is always down, regardless of whether something is moving up or down. Whether you call up positive depends on your convention--you can always choose up as positive, which makes 'a' negative (for free fall type problems).

For example for this question, i got it wrong =(

A projectile is launched straight up at 60.0m/s from a height of 80m at the edge of a cliff. The projectile falls, just missing the cliff and hitting the ground below. Find a) max height of the projectile above the point of firing.

In order to find the max height, i calculated the time it took to get to the max height, where v = 0. this is what i did:

v = vo +at
0 = 60 + 9.8t
t = -9.18s

No. Using the convention, the initial velocity would be + 60 m/s (since it's going up) and the acceleration would be -9.8 m/s^2 (since gravity acts down).

When you get a negative time, that's a clue that something went wrong!

then i solved for the max height using the equation: y = voyt + 1/2at^2

is that correct??

Nothing wrong with that approach, once you have the correct signs.

But why didn't you just use the equation you gave before? Namely:
v^2 = vo^2 + 2a(x2-x1)

That would give you the answer in one step. Give that equation a try and you can compare answers.