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

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A brick is thrown upward from a building at an angle of 23.8 degrees with an initial speed of 14.2 m/s, and it remains in flight for 3.3 seconds. The kinematic equation used is d = vi*t + 1/2*a*t^2, where gravity is 9.8 m/s². The vertical component of the initial velocity was calculated to be 5.73 m/s, leading to a displacement calculation that resulted in 24.899 meters. However, the initial attempt was marked incorrect due to a miswritten kinematic equation and incorrect final velocities. The discussion confirms that the vertical displacement does indeed represent the height of the building.
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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