Please help me. topic: ( motion with constant acceleration : free fall )

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

The Attempt at a Solution

Homework Statement

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

Homework Equations

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 !

 

[Doc Al]

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 + volt + 1/2(-g)T^2

Use this formula to solve for the time.

 

[donniemsb_12]

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]

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?

 

[donniemsb_12]

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]

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.)

 

[donniemsb_12]

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]

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.

 

[donniemsb_12]

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]

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

Yes.

 

[donniemsb_12]

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]

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?

 

[donniemsb_12]

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]

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.)