Building Height Calculated - Thanks Thaakisfox!

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Homework Help Overview

The discussion revolves around calculating the height of a building from which a ball is thrown horizontally. The problem involves horizontal motion and free fall, with specific values provided for initial velocity and horizontal distance traveled.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to calculate the time of descent based on horizontal displacement and velocity. There are questions about the appropriate formulas to use for horizontal projection and free fall, with some confusion regarding the application of trigonometric functions.

Discussion Status

Participants are actively engaging with the problem, exploring different interpretations of the motion involved. Some guidance has been offered regarding the calculations, and there is recognition of potential issues with the problem's parameters, though no consensus has been reached on the final interpretation.

Contextual Notes

There is a mention of the given data possibly being unrealistic, as it suggests a building height that exceeds typical structures. This raises questions about the validity of the problem setup.

TeenieWeenie
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Problem solved! Thanks again Thaakisfox!

Homework Statement


A ball thrown horizontally at 2.2 m/s from the roof of a building lands 36 m from the base of the building. Calculate the height of the building.
Vox = 2.2 m/s
Xo = 0m
Vx = 0 m/s
X = 36m

Homework Equations


y = Vosin ao - (0.5)(g)(t^2)

The Attempt at a Solution


I tried plugging it in but then I have a missing time...?
 
Last edited:
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This is a horizontal projection hence you don't need any sine etc. Just calculate the time of descent from the velocity and horizontal displacement by dividing them. And then the vertical component is simple freefall.
 
I'm kinda confused.
What formula would I use then?

x=Vo *CosAo * t
 
No. It is a horizontal projection. You don't need any of those sine or cosine function.

You throw the ball with horizontal velocity Vo. Since it has no acceleration in the horizontal direction, the distance it travels is simply: x=Vo*t where x is given (the distance from the base of the building). from here you can get t. Now in the vertical plane it is just simple freefall, so use the formula which gives the distance when freefall takes place
 
36 m = 2.2 m/s * t

36/2.2 s = t
16.36 s = t

Formula for free fall: h(t) = Vo*t + 1/2 at^2
h(16.36) = 0*t + (0.5)(-9.8m/s^2)(16.36^2)
= -1311.48304 m ?

I think something went wrong... :(
 
The calculation is correct. The given data seem insensible.
 
Insensible?
 
That happens many times in textbooks, that the given data for a problem don't make sense. For example there probably arent many buildings taller than 1km, and especially someone throwing a ball off.
But your calculation is correct. (That minus sign doesn't matter, it just means that you took the y-axis to point upwards, and that's why the acceleration has a negative sign. But you are searching for the absolute value of the height anyway, so just take that minus sign away.)
 
Oh! Good observation.

Problem solved! Thanks again Thaakisfox!
 

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