Physics Conceptual Question

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MedPR

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This is a homework problem, so if you don't want to read the thread I understand.

I don't remember the exact numbers, but I came across this projectile motion problem while doing homework and I thought the solution in the solutions manual was wrong, but my teacher assured me it is correct.

The problem goes like this:

While standing on top of a building, you launch a brick in a direction theta above the horizontal. The flight time is 3 seconds. How high is the building?

I bolded building for emphasis.

The solution's manual says you can simply plug in the given values to y=voyt-1/2gt^2 and the answer, y, is the height of the building.

To me it seems like the y in that equation would be the max height that the brick reaches above the ground, which is taller than the building itself.

I asked my teacher about it and he said "The equation accounts for the additional height. It is a very common mistake for students to think that they need to break this problem up into two parts, the height of the projectile, and the height of the building"

I'm still confident that I am right in thinking that you need to find the time that the brick spends above the building, subtract that time from 3seconds (given as the total flight time), then use that time in the equation y=voyt-1/2gt^2.

If I'm wrong, can someone explain what I am missing?

TieuBachHo

Full Member
10+ Year Member
7+ Year Member
This is a homework problem, so if you don't want to read the thread I understand.

I don't remember the exact numbers, but I came across this projectile motion problem while doing homework and I thought the solution in the solutions manual was wrong, but my teacher assured me it is correct.

The problem goes like this:

I bolded building for emphasis.

The solution's manual says you can simply plug in the given values to y=voyt-1/2gt^2 and the answer, y, is the height of the building.

To me it seems like the y in that equation would be the max height that the brick reaches above the ground, which is taller than the building itself.

I asked my teacher about it and he said "The equation accounts for the additional height. It is a very common mistake for students to think that they need to break this problem up into two parts, the height of the projectile, and the height of the building"

I'm still confident that I am right in thinking that you need to find the time that the brick spends above the building, subtract that time from 3seconds (given as the total flight time), then use that time in the equation y=voyt-1/2gt^2.

If I'm wrong, can someone explain what I am missing?
Your teacher is right. If you are still unsure, you can look it up somewhere for the equation. That's the pitfall if you are not careful. Y is the height at launch to where the object lands. The maximal height occurs when voy = 0.

milski

1K member
5+ Year Member
The solution is correct. v0 is the initial vertical velocity, y on the left side is the displacement. You'll get it to be 0 (same as top of the building) at moment t=0 and a bit later, at t=x.xxx, when the brick is falling down. The final position of the brick at t=3 sec will be -y.yyy and that will be the height of the building, since you chose the top of the building as an origin of your coordinate system.

Max height will be when v=0, v0 is a constant.

MedPR

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I see. I think I understand now.

The fact that voy isn't 0 makes it work.