Why Does a Free-Falling Object Not Have Zero Final Velocity Upon Impact?

  • #1
Ineedhelpwithphysics
42
7
Homework Statement
When I drop a rock from the top of my ladder, I notice that it takes 0.25 seconds for the rock to fall past our 2-m tall door and hit the ground below. What was the velocity of the rock when it passed the top of the door?
Relevant Equations
vf = v0 + at
∆x = v0t + 1/2 at^2
I am super stumped at this question, the answer key is telling me 6.78 downwards, i think I'm reading and observing the question wrong. Isn't final velocity 0 so why can't i do

0 = v0 -9.8(0.25)
-2.45 = v0

But when I use the second the equation
-2 (displacement of door) = v0(0.25) + 1/2 (-9.8)(0.25)

-6.75 = v0

Why am i getting two answers, it's clear the first equation is wrong but why is that? is the final velocity not zero, but it hit the ground and hitting the ground means 0 velocity.
 
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  • #2
Ineedhelpwithphysics said:
Homework Statement: When I drop a rock from the top of my ladder, I notice that it takes 0.25 seconds for the rock to fall past our 2-m tall door and hit the ground below. What was the velocity of the rock when it passed the top of the door?
Relevant Equations: vf = v0 + at
∆x = v0t + 1/2 at^2

is the final velocity not zero, but it hit the ground and hitting the ground means 0 velocity.
Yes the final velocity (right before the event of hitting the ground) is not zero. Kinematics don't study collisions at least not in depth, so you should know that by final velocity here we don't consider the velocity after hitting the ground, but the velocity right before it hits the ground.
 
  • #3
Ineedhelpwithphysics said:
Homework Statement: When I drop a rock from the top of my ladder, I notice that it takes 0.25 seconds for the rock to fall past our 2-m tall door and hit the ground below. What was the velocity of the rock when it passed the top of the door?
Relevant Equations: vf = v0 + at
∆x = v0t + 1/2 at^2

I am super stumped at this question, the answer key is telling me 6.78 downwards, i think I'm reading and observing the question wrong. Isn't final velocity 0 so why can't i do

0 = v0 -9.8(0.25)
-2.45 = v0

But when I use the second the equation
-2 (displacement of door) = v0(0.25) + 1/2 (-9.8)(0.25)

-6.75 = v0

Why am i getting two answers, it's clear the first equation is wrong but why is that? is the final velocity not zero, but it hit the ground and hitting the ground means 0 velocity.

I'm kinda' slow sometimes, but unless you tell us the height of the ladder where the object was dropped, it's kind of hard to answer this question, no?
 
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  • #4
He says 2m tall door but not sure if he correct -edit it right now. Ah yes i guess it is implied height of ladder=height of door.
 
  • #5
berkeman said:
I'm kinda' slow sometimes, but unless you tell us the height of the ladder where the object was dropped, it's kind of hard to answer this question, no?
it's not given it asks the height of the ladder after the first question.
 
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  • #6
Ineedhelpwithphysics said:
it's not given it asks the height of the ladder after the first question.
WITW? What is the full Problem Statement?
 
  • #7
oops the problem is abit more complex than what I thought.

Right now the full picture as I get it is that the ladder is a bit higher than the door, we let the rock fall with zero initial velocity and then (after time ##t_0##) the rock passes the top of the door, and after ##2.5sec=\Delta t## it is at the bottom of the door.
 
  • #8
berkeman said:
WITW? What is the full Problem Statement?
When I drop a rock from the top of my ladder, I notice that it takes 0.25 seconds for the rock to fall past our
2-m tall door and hit the ground below. What was the velocity of the rock when it passed the top of the door?
How tall is my ladder?

You're suppose to take -6.78 m/s as the final velocity, the initial velocity is 0, acceleration is -9.8 m/s . Im assuming you use those values for this equation ∆x = v0t + 1/2 at^2. That is the displacement from the top of the door to the top of the ladder. You then add 2m + 2.3 m (displacement between the top of the door and top of the ladder) and you get 4.3m.

Answer key says: -6.78 m/s velocity when it passed the top of the door) and 4.3m (meter of the ladder) .
 
  • #9
Ineedhelpwithphysics said:
You're suppose to take -6.78 m/s as the final velocity, the initial velocity is 0, acceleration is -9.8 m/s . Im assuming you use those values for this equation ∆x = v0t + 1/2 at^2. That is the displacement from the top of the door to the top of the ladder. You then add 2m + 2.3 m (displacement between the top of the door and top of the ladder) and you get 4.3m.
You have to use the equation $$\Delta x=v_0 \Delta t+\frac{1}{2} g(\Delta t)^2$$ in two cases: One case is for the displacement from the top of door to the bottom of door (ground) and the other case for the displacement from the top of ladder to the top of door.

You did correctly the first case where you calculate ##v_0##=the velocity the rock has when it passes the top of the door=6.75m/s

But you don't tell us what you did exactly for the second case, that is how you calculate the displacement between the top of the door and the top of the ladder.
 
  • #10
Ehm now that I read the problem statement again, I feel an ambiguity on what exactly the 0.25sec time interval is for.. Is it for the time from the top of ladder till it hits the ground, for the time from the top of ladder to the top of door, or for the time from the top of door to the bottom of door-ground?
 
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  • #11
If I interpreted the problem correctly (as I state in post #7) and with g=10m/s^2 the answers I get is ##v_0=6.75## and ##h=4.27m## h being the height of ladder.
 
  • #12
Delta2 said:
Ehm now that I read the problem statement again, I feel an ambiguity on what exactly the 0.25sec time interval is for.. Is it for the time from the top of ladder till it hits the ground, for the time from the top of ladder to the top of door, or for the time from the top of door to the bottom of door-ground?
I think i get it now.

6.78 is the initial velocity at the top of the door
To find the height of the ladder I would have to make 6.78 the final velocity, and the initial velocity 0
I find the time between the rock dropping from the top of the ladder to the top of the door.
Then using the kinematics equations I would find the displacement from the top of the door to the beginning of the ladder.
Then i would add that by 2.
 
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  • #13
Delta2 said:
oops the problem is abit more complex than what I thought.

Right now the full picture as I get it is that the ladder is a bit higher than the door, we let the rock fall with zero initial velocity and then (after time ##t_0##) the rock passes the top of the door, and after ##2.5sec=\Delta t## it is at the bottom of the door.
No. the problem is badly stated. The only way this will work out is that the ladder is higher than the door and the 0.25s (not 2.5s) is the time interval from door top passage to ground impact. So the rock must be at nearly 8 m/s when it passes the top of the door. Find that number exactly and then you know the height of the ladder above the door. Done. Let us know the result. My (±10%) estimate is the ladder is about 5.2 m from floor
 
  • #14
hutchphd said:
No. the problem is badly stated. The only way this will work out is that the ladder is higher than the door and the 0.25s (not 2.5s) is the time interval from door top passage to ground impact. So the rock must be at nearly 8 m/s when it passes the top of the door. Find that number exactly and then you know the height of the ladder above the door. Done. Let us know the result. My (±10%) estimate is the ladder is about 5.2 m from floor
How did you find 8m/s there I find 6.75m/s (use g=10m/s^2).
 
  • #15
Because it takes .25s to travel 2m.
REV :During .25s it will add only 2.5 m/s so indeed at the top it will be 8-1.25=6.75m/s. My estimate was worse than i thought. Never do late night physics when you are old!
 
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  • #16
Ineedhelpwithphysics said:
Isn't final velocity 0
The "SUVAT" equations you are using are for the special case of kinematics in which acceleration is constant. When it hits the ground, the acceleration suddenly changes from g downwards to something huge upwards.
 
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  • #17
hutchphd said:
Because it takes .25s to travel 2m.
Yes right, I used that too , full equation $$2=v_0(0.25)+0.5g (0.25^2)$$
 
  • #18
Delta2 said:
How did you find 8m/s there I find 6.75m/s (use g=10m/s^2).
The 8m/s was a ballpark number, not intended to be accurate.
 
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  • #19
Your number is correct! My estimate was worse than intended and indicated (see my revision)
.
 
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  • #20
It stumps me that this question has generated such an extensive thread. There really is not much more to it than writing down the
$$
\Delta x = v_0 t + \frac{gt^2}{2}
$$
of the OP and solving for ##v_0##. For the height of the ladder (above the door), just find the time of zero velocity and insert it into the equation above.
 
  • #21
Orodruin said:
It stumps me that this question has generated such an extensive thread. There really is not much more to it than writing down the
$$
\Delta x = v_0 t + \frac{gt^2}{2}
$$
of the OP and solving for ##v_0##. For the height of the ladder (above the door), just find the time of zero velocity and insert it into the equation above.
There was some confusion regarding which events were separated by 0.25s. The wording is unclear.
 
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  • #22
haruspex said:
There was some confusion regarding which events were separated by 0.25s. The wording is unclear.
The statement in the OP seems pretty clear to me. If it takes 0.25 s to fall past a 2 m door, that would be the time from the top of the door to the bottom of the door. That would seem to leave little room for interpretation to me.
 
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  • #23
Ineedhelpwithphysics said:
is the final velocity not zero, but it hit the ground and hitting the ground means 0 velocity.
If you drop a rock from ##1m##, then it's velocity when it hits the ground is about ##4.5m/s##. And it takes about ##0.45s## to fall. That's free-fall motion. What it does after it hits the ground is a different matter. It might bounce up, sideways or shatter into several pieces that fly off in different directions. All that stuff is not captured by the equation ##v = gt##. How could it be?
 
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