still, no one has been able to answer this correctly?

Question

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)

Answer 1

The building is 19.23m tall.

Explanation:

The only trick in this problem is knowing that the only part of the initial velocity needed is the vertical component. This is the initial velocity times the sin of the projection angle (18.9 degrees):

Viy = Vi * sin(18.9) = 6.87 m/s

Using this value, plug everything you know into the acceleration formula:

0 = -0.5*g*t^2 + Viy*t + h

Where g is the acceleration due to gravity, t is the time in flight, Viy is the vertical component of the initial velocity and h is the height of the building (the initial height).

Solve this equation for h to get:

h = 0.5*g*t^2 - Viy*t = 0.5*9.81*2.8^2 - 6.87*2.8 = 19.2m

Good luck with your physics, I hope this helps!

Answer 2

You know the initial velocity (21.2 m/s), so get the vertical component by using the sine of the given angle (18.9 degrees). Then note that the horizontal and vertical motions are independent, so it becomes a constant acceleration problem, with a given initial velocity. Solve for the time until the final velocity equals 0. Then subtract twice that time from the 2.8 s. That way, you subtract the time the object is above the initial thrown level. Now move to the second part of the problem.

The motion is symmetrical in the vertical, so the initial velocity at the level of the building will be the same as at the start, but downward. You know the acceleration due to gravity (9.8 m/s^2) downward. You also know the time the brick is falling. Use these facts to calculate the distance the brick falls, which will be the height of the building (or the height from which the brick is thrown).

The height from which the brick is thrown is 19.19 meters above the surrounding surface. The brick impacts at a velocity of 20.57 m/s downward.

All of this neglects the effects of friction (air or otherwise) on the brick, any angular misalignment caused by wind "drift" to introduce another plane into the equation, and that the projectile is actually launched from "ground zero" on the rooftop ... not at the rooftop plus some launcher's height).

Answer 3

You know the velocity, so get the vertical component of the initial velocity by using the sine of the given angle. Then note that the horizontal and vertical motions are independent, so it becomes a constant acceleration problem, with a given initial velocity. Solve for the time until the final velocity equals 0. Then subtract twice that time from the 2.8 s. That way, you subtract the time the object is above the initial thrown level. Now move to the second part of the problem.

The motion is symmetrical in the vertical, so the initial velocity at the level of the building will be the same as at the start, but downward. You know the acceleration is 9.8 m/s^2 downward. You also know the time the brick is falling. Use these facts to calculate the distance the brick falls, which will be the height of the building (or the height from which the brick is thrown).

The height from which the brick is thrown is 19.19 meters above the surrounding surface. The brick impacts at a velocity of 20.57 m/s downward.

Answer 4

First, you figure out the vertical component of the initial velocity:
v0(vertical) = 21.2 m/s * sin(18.9 degrees) = 6.87 m/s
then you use the formula
x = v0 * t + 1/2 * a * t^2
You need to make two definitions. First, define the point from which the brick is thrown as 0 on the x scale, and define up as being in the positive direction; therefore, the initial velocity you just calculated is positive, while gravity is negative
x = (6.87m/s) * (2.8s) + 1/2 (-9.8m/s^2) * (2.8s)^2
Plug the numbers into your calculator, and you'll find that x is -19.19 m...in other words, the brick ends up 19.19 m below the point from which it was thrown.

Answer 5

The initial speed is 21.2 m/s. The fraction of this speed at which the brick is rising is equal to the sine of 18.9 degrees, which is 0.3239. So it starts with an initial vertical velocity of 0.3239 * 21.2 m/s, or 6.867 m/s. The velocity decreases at a rate of 9.8 m/s^2 until v = 0. This process takes (6.867/9.8), or 0.701, s, and during this time the average velocity of the brick is (6.867/2) or 3.436 m/s, so the brick rises (3.436 * 0.701) or 2.409 meters.

The brick then begins to fall, starting with a velocity of 0 and accelerating at 9.8 m/s until it hits the ground. This process takes (2.800-0.701) or 2.099 seconds and in this time the brick accelerates to (2.099 * 9.8) or 20.570 m/s. The brick's average velocity is half this amount, or 10.285 m/s. Therefore, in the time between when it begins to fall and when it hits the ground, the brick travels (10.285 * 2.099) or 21.588 meters.

But remember, it was higher than the top of the building when it started to fall. The height of the building is the distance it fell (21.588 m) minus the distance it rose (2.409 m), or 19.179 m.

Answer 6

Well, I could possibly do that...but I'm far too lazy. Also, you did not tell whether we were assuming that it was within a vaccuum, or if we were to consider air resistance. Lastly, and this is completely off topic, but that's got to be one strong person to throw a brick that fast. Sorry I couldn't help you. Try asking some where that won't have as many lazy people.

Answer 7

Why insist upon asking a question that you know the answer to, to prove how intelligent you are capable to get the answer, while many of us are not able to come up with the answer.

Answer 8

well its mechanics motion mechanics i can easly do that but why the mind hurt while i can get 2 points clean with that answer ; ) u can keep the other 8 points i just want 2