A brick is thrown upward from the top of a building at an angle of 30 degrees to the horizontal and with an initial speed of 13 m/s. If the brick is in flight for 2.9 s, how tall is the building?

Question

anonymous · Answer

okay i have worked it not sure if im correct. i used the following for the answer i got 
1: $$h=u \times \sin \alpha \times t- \frac{ g }{ 2 }t ^{2}$$
u=initial velocity=13m/s
$$\alpha=30^{0}$$
t=2.9s
$$g=9.8m/s ^{2}$$
once i had all my values i just plugged it into the formula and i got an answer.
the only thing is you are going to get a negative answer because the horizontal plane at which the brick left the top of the building is higher than the building itself so the answer you get is negative because the brick went lower then its starting point so the negative number is the height it fell below the original horizontal plane. which means any distance it fell lower then the horizontal plane means it is the height of the building. or you could say the distance it fell is was along the height of the building.

anonymous · Answer

okay does the formula say it is divided by g or is it a subtraction sign? I just can't make it out.

anonymous · Answer

its a subtraction