How Tall Is the Building If a Brick Is Thrown Upward?

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Homework Help Overview

The problem involves a brick thrown upward from the top of a building at an angle, with given initial speed and time of flight. The subject area includes kinematics and projectile motion, specifically focusing on vertical displacement and its relation to the height of the building.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of trigonometry to resolve initial velocities into vertical and horizontal components. There is an attempt to apply kinematic equations to find the height of the building based on vertical displacement. Questions arise regarding the correctness of the kinematic equation used and the interpretation of vertical displacement as the building's height.

Discussion Status

Some participants have provided guidance on checking the kinematic equation and the calculations involved. There is a recognition of the need to clarify the relationship between vertical displacement and the height of the building, with some participants affirming the interpretation of vertical displacement as the height.

Contextual Notes

There is mention of an online assignment indicating an incorrect answer, prompting further examination of the calculations and assumptions made regarding vertical displacement and kinematic equations.

MDT5507
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Homework Statement


A brick is thrown upward from the top of a building at an angle of 23.8 degrees above the horizontal and with an initial speed of a 14.2 m/s. Acceleration of gravity is 9.8 m/s/s. If the brick is in flight for 3.3 s how tall is the building?

Homework Equations


d=vi*t+1/2*a*t^2, trigonometry (sin=opp/hyp cos=adj/hyp)

The Attempt at a Solution


I used trig to get my vertical and horizontal vf and vi(horizontal 12.99 & 12.99, vertical 5.73 & -5.73) Then i plugged in these and my time into the kinematic equation. d=5.73*3.3+-4.9+3.3^2. My d in this answer would be the height of the building I thought, but the online assignment says its incorrect. I believe the vertical displacement would be the height of the building and i solved it to be 24.899.
 
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MDT5507 said:

Homework Statement


A brick is thrown upward from the top of a building at an angle of 23.8 degrees above the horizontal and with an initial speed of a 14.2 m/s. Acceleration of gravity is 9.8 m/s/s. If the brick is in flight for 3.3 s how tall is the building?

Homework Equations


d=vi*t+1/2*a*t^2, trigonometry (sin=opp/hyp cos=adj/hyp)

The Attempt at a Solution


I used trig to get my vertical and horizontal vf and vi(horizontal 12.99 & 12.99, vertical 5.73 & -5.73) Then i plugged in these and my time into the kinematic equation. d=5.73*3.3+-4.9+3.3^2. My d in this answer would be the height of the building I thought, but the online assignment says its incorrect. I believe the vertical displacement would be the height of the building and i solved it to be 24.899.
Check your kinematic equation again ...you wrote it down incorrectly. Your approach is ok, but the final velocities when the brick hits the ground are incorrect, but not needed here.
 
Thanks, so am I correct that the vertical displacement would give me the height of the building?
 
MDT5507 said:
Thanks, so am I correct that the vertical displacement would give me the height of the building
Yes, correct.
 

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