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

[MDT5507]

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.

 

[PhanthomJay]

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.

 

[MDT5507]

Thanks, so am I correct that the vertical displacement would give me the height of the building?

 

[PhanthomJay]

Thanks, so am I correct that the vertical displacement would give me the height of the building

Yes, correct.

 

[HalfLostAndSearching]

Your attempt at a solution is correct! The vertical displacement of the brick is indeed equal to the height of the building. However, there may be a small error in your calculation. When plugging in the values into the kinematic equation, be sure to use the correct signs for the acceleration and the time. In this case, the acceleration should be negative (due to gravity pulling the brick downwards) and the time should be positive. Also, be sure to use the correct units for all values (meters and seconds in this case). So the correct calculation would be:

d = (5.73 m/s)(3.3 s) + (1/2)(-9.8 m/s^2)(3.3 s)^2
= 18.9 m + (-52.7 m)
= -33.8 m

The negative sign indicates that the displacement is in the downward direction, which makes sense since the brick is thrown upwards and then falls back down. So the height of the building is actually 33.8 m.