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

AI Thread Summary
A man drops a can from a scaffolding rising at 1.0 m/s from a height of 50.0 m. To determine the time for the can to reach the ground and its final velocity, the relevant equations of motion under constant acceleration due to gravity (g = -9.8 m/s²) are discussed. The correct approach involves using the formula y = yo + Vot + 1/2(-g)t² to solve for time and final velocity. Confusion arises regarding the sign of g and the need to disregard an illustration provided by the professor. Ultimately, the focus should be on applying the correct formulas to find the solutions in a single computation.
donniemsb_12
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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)

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 !
 

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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 ?
 
  • #10
donniemsb_12 said:
so sir. i will be focus on the formula that u was given to me ?
Yes.
 
  • #11
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 !
 
  • #12
donniemsb_12 said:
t^2=-2(y)/g
t^2=2(50)/9.8

t=3.19 s
:confused: Why didn't you use the formula I indicated in post #2?
 
  • #13
Doc Al said:
:confused: 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 .
 
  • #14
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.)
 
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