Velocity-time graph

1. Mar 27, 2012

azizlwl

1. The problem statement, all variables and given/known data
The velocity–time graph for the vertical component of the velocity of an object
thrown upward from the ground which reaches the roof of a building and
returns to the ground is shown in below. Calculate the height of the building.
http://img832.imageshack.us/img832/9528/roofr.jpg [Broken]

2. Relevant equations
s=vt/2

3. The attempt at a solution
What I am thinking is that the height of the roof must be at the highest reach of the object ie. when velocity is zero. Then the total must be the total area of positive side of velocity-time graph.

1. The problem statement, all variables and given/known data
h =1/2 × 3 × 30 −1/2 × 1 × 10 = 40m

Using above solution, it means the object goes higher than the roof.

Last edited by a moderator: May 5, 2017
2. Mar 27, 2012

spacelike

The whole question and solution seems a bit odd to me.

First because if the object fell back to its starting point then the graph should go down to -30m/s, but instead the graph shows it as stopping at -10m/s, I was assuming that it just went below the graph but they just don't show it, but then the books solution uses the -10m/s so that is strange.

Also, the book is finding the total displacement of the object, because they are including the negative side.
So I am wondering, are you sure that the question isn't that an object is thrown up and lands on top of the roof of a building?

Because in that case their answer makes sense. It would have to go above the building, and then fall down a little to land on top of it, where the total area (including the negative side) would give the displacement of the object, and equivalently the height of the building.

3. Mar 27, 2012

azizlwl

This is the exact answer from the book.

Height h = area under the υ − t graph. Area above the t-axis is taken positive
and below the t-axis is taken negative. h = area of bigger triangle minus area
of smaller triangle.
Now the area of a triangle = base × altitude
h =1/2 × 3 × 30 −1/2 × 1 × 10 = 40m

4. Mar 27, 2012

spacelike

Right, but what I was asking is if you mistyped/misunderstood the question,

There are 3 parts to your original post: question, graph, books answer.
2 of them work with each other, the graph and the books answer
the thing that doesn't fit with those two is the question.

So I suspect that the book question is actually saying that the ball comes to rest on top of the roof of the building.
Can you verify this?

5. Mar 27, 2012

azizlwl

http://img585.imageshack.us/img585/5615/graphic1gv.jpg [Broken]

Last edited by a moderator: May 5, 2017
6. Mar 27, 2012

spacelike

Well something is definitely wrong.

Either it is an error in the book, or a very poorly worded question that both of us are just misunderstanding somehow.. (although it seems pretty clear).
If the graph showed the entire trajectory it should continue going down to -30m/s

There is just a lot of things wrong here.. Seems like a pretty messed up question, tell me if I'm misunderstanding:

1) The books answer suggests that the ball goes above the building.
2) The question says that the ball falls all the way back down to the floor.
3) The question suggests that the balls maximum height is the building which contradicts (1).
4) The graph suggests that the ball doesn't fall all the way back down to the floor which contradicts (2).

NOTE: All of that could be resolved if the ball goes above the roof and lands on top of the building. But the question clearly states that it falls to the floor...
So I don't know what to say.. It seems to me there is an error somewhere in the book (question, graph, answer, somewhere).

Sorry, it happens.

7. Mar 27, 2012

azizlwl

Thanks spacelike. Since English is not my first language, i guess i've made mistake in understanding the question. So we may conclude the diagram may be wrong. Even the x-intercept is not exactly at 3 which is easily visible.