A brick is thrown upwards from the top of a building at an angle of 18.9 degrees above the horizontal and with an initial speed of 21.2 m/s. (The acceleration due to gravity is 9.8 m/s^2). If the brick is in flight for 2.8 s, how tall is the building? (in meters)
2006-11-14 08:40:55 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Physics
I can't believe no one has answered. It's so obvious. Let's give 'em a little more time.
You and I are so smart.
2006-11-14 08:46:08 · answer #1 · answered by faversham 5 · 1⤊ 2⤋
the distance=ut+1/2gt^2
=21.2(2.8)-1/2*9.8(2.8)^2
=59.36-38.42=21 m
so the height=21m approx.
2006-11-14 16:49:32 · answer #2 · answered by raj 7 · 0⤊ 1⤋
v0 = 21.2 * sin(18.9) = 6.867 m/s
g = -9.8 m/s^2
x(2.9) = 0 = 0.5 * g * t^2 + v0 *t + x0 = -21.3 + x0
x0 = 21.3 m
2006-11-14 16:46:44 · answer #3 · answered by feanor 7 · 0⤊ 0⤋
Its about 21 meters tall. I did it in my head, so don't expect me to show my work
2006-11-14 16:48:27 · answer #4 · answered by Anonymous · 0⤊ 2⤋
Assume no air drag.
...... ^ .................... ^
.... /...\ ................. /..\
Vo/.....\.......=...Vo/....\V1...+...\V1
|..........\....................................\
h..........\....................................\
|............\Vf.................................\Vf
...-dtotal-.............--d1--.........-d2-
Just doodling here's the answer:
At the topof the perabola v=0=Vy-at=>
t to top of perabola = Vy/a=21.2sin(18.9)/9.8=0.700719313s
Vx=21.2cos(18.9) d=Vt=>
d1=Vx*2(0.700719313)=28.10866798m
dtotal=Vx*(2.8s)=56.1596269m
d2=28.05095892m
Don't know why I looked for distances... Hmmm
|Vo|=|V1| => V1y=Vy=21.2sin(18.9)
h=height of building
h=V1y+a*t=21.2sin(18.9)+9.8*(2.8-2*0.700719313)
height of building = 32.93m
Is that right or am I blonde?
2006-11-14 17:35:49 · answer #5 · answered by unlv_engineer 2 · 0⤊ 1⤋
Hey! do you go to Havard?
2006-11-14 16:57:40 · answer #6 · answered by Simple1 6 · 0⤊ 1⤋