## Homework Statement

A scaffolding is rising at 1.0 m/s when at a height of 50.0 m , then the man on the scaffolding drop a can. determine:

a. the time needed for the can to reach the ground .

b. what is the final velocity ?

## Homework Equations

note: g=-9.8 m/s
the letter o in the equation stands for initial example : Vo( initial velocity) .

V=Vo +(-g)t
y=yo + Vot + 1/2(-g)T^2
V^2=Vo^2 + 2(-g)(y-yo)
y-yo=(v+vo)/2 (t)

t = time
y=position
v = velocity
g=gravity

## The Attempt at a Solution

The illustration was illustrated by our professor .

the attempt that i was trying to solve is i tried to get each of all the final velocity.

guys can you help me on how to solve this ! thank you !

#### Attachments

• illustration.jpg
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Doc Al
Mentor
note: g=-9.8 m/s
That's confusing. I assume you mean that a = -g, where g = 9.8 m/s^2.

y=yo + Vot + 1/2(-g)T^2
Use this formula to solve for the time.

That's confusing. I assume you mean that a = -g, where g = 9.8 m/s^2.

Use this formula to solve for the time.

sir yes ! a= g . and the g is equals to -9.8 m/s^2

sir ! how many times should i compute for the time ? is it 1 time or 4 times ? base on the illustration ???

Doc Al
Mentor
sir yes ! a= g . and the g is equals to -9.8 m/s^2
If you insist on g being negative, then your formula is incorrect. (It has a -g in it.)
sir ! how many times should i compute for the time ? is it 1 time or 4 times ? base on the illustration ???
Why not just once? Solve for the final time when it hits the ground. What values will you put for y, y0, and v0?

If you insist on g being negative, then your formula is incorrect. (It has a -g in it.)

Why not just once? Solve for the final time when it hits the ground. What values will you put for y, y0, and v0?

sir ! our professor says that it is always negative . so he is wrong ? and the right is -g=9.8 m/s^2 ? right ? thank you ! so sir we will disregards the illustration ?? and compute the final velocity just once ?

Doc Al
Mentor
sir ! our professor says that it is always negative . so he is wrong ? and the right is -g=9.8 m/s^2 ? right ?
Taking up as positive, the acceleration is -9.8 m/s^2.

(The constant g is usually taken as positive, so the acceleration would be -g.)

Taking up as positive, the acceleration is -9.8 m/s^2.

(The constant g is usually taken as positive, so the acceleration would be -g.)

ah i see sir ! thank you. sir so i will disregard the illustration that our prof was made ? and i will focus in solving using the given value's in the problem ?

Doc Al
Mentor
ah i see sir ! thank you. sir so i will disregard the illustration that our prof was made ? and i will focus in solving using the given value's in the problem ?
I suppose the illustration is there to help you visualize what's going on. (I think the velocity at point 3 should be negative.) But you can solve for the final time in one step, you don't need to solve for all the intermediate points.

I suppose the illustration is there to help you visualize what's going on. (I think the velocity at point 3 should be negative.) But you can solve for the final time in one step, you don't need to solve for all the intermediate points.

so sir. i will be focus on the formula that u was given to me ?

Doc Al
Mentor
so sir. i will be focus on the formula that u was given to me ?
Yes.

Yes.

t^2=-2(y)/g
t^2=2(50)/9.8

t=3.19 s

y-50=v^2-(1.0)^2/2(9.8)
v^2-(1.0)^2 = (-50)*2(9.8)
v^2-1 =-980
V^2/-1=-980/-1
v^2=980

V=31.30 m/s

sir am i right ? regards on solving the problem ? thank you !

Doc Al
Mentor
t^2=-2(y)/g
t^2=2(50)/9.8

t=3.19 s Why didn't you use the formula I indicated in post #2? Why didn't you use the formula I indicated in post #2?

sir i derived the formula that you indicate like what i saw in my lecture .

Doc Al
Mentor
sir i derived the formula that you indicate like what i saw in my lecture .
You posted some formulas. I quoted one and said "use this one". So use it. (The one you derived assumes that the initial speed is zero.)