• donniemsb_12
In summary: 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.)
donniemsb_12

## 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
13.8 KB · Views: 469
donniemsb_12 said:
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.

Doc Al said:
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 ?

donniemsb_12 said:
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?

Doc Al said:
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 ?

donniemsb_12 said:
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.)

Doc Al said:
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 ?

donniemsb_12 said:
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.

Doc Al said:
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 ?

donniemsb_12 said:
so sir. i will be focus on the formula that u was given to me ?
Yes.

Doc Al said:
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 !

donniemsb_12 said:
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?

Doc Al said:
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 .

donniemsb_12 said:
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.)

## What is free fall?

Free fall is a type of motion where an object falls freely under the influence of gravity, without any other forces acting on it.

## What is constant acceleration?

Constant acceleration is when an object's velocity changes by the same amount in each unit of time. In free fall, the acceleration due to gravity is considered constant.

## How is the motion of an object in free fall described?

The motion of an object in free fall is described by the laws of motion and the equation of motion, which states that the displacement of the object is equal to its initial velocity multiplied by time, plus half the acceleration due to gravity multiplied by the square of time.

## What factors affect the motion of an object in free fall?

The motion of an object in free fall is affected by the mass of the object, the acceleration due to gravity, and the air resistance or drag on the object.

## How is the acceleration due to gravity related to free fall?

The acceleration due to gravity is the force that pulls objects towards the center of the Earth. In free fall, this acceleration is considered constant and is equal to 9.8 meters per second squared (m/s²) near the surface of the Earth.

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