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

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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?