Projectile Motion?

Question

A brick is thrown upward from the top of a building at an angle of 20° to the horizontal and with an initial speed of 11 m/s. If the brick is in flight for 3.3 s, how tall is the building?

can you help me but not tell me the answer? i know it deals with trigonometry but i don't know how to set up the equation to solve it.

Answer 1

The height of the brick from the release point is

h = Vy *t - (g/2)*t^2 = -(the building height) when t = 3.3sec

Vy is 11sin20

Plug 'n play........

Answer 2

Sure. The problem gives you a vector : 11m/s at 20 degrees above the horizontal. Convert this into a component vector by using the following equations:

x component = 11 m/s * cos(20)
y component = 11 m/s * sin(20)

You'll get a vector with an x and a y component. Now, use the three projectile motion equations with the y component to figure out the original y position. Remember, the ending y position is 0, and the time is 3.3 seconds. The starting velocity is whatever your y component came out to be. Acceleration is 9.8 m/s^2.

Just make sure to keep your signs consistent: keeping up as positive is usually easier. Using up as positive, your starting y velocity would be positive, and acceleration is negative 9.8 m/s^2. Just plug into the equation and solve.

Answer 3

OK, consider the horizontal motion and vertical motion separately. You don't really care about the horizontal motion. The initial vertical speed upward is 11 m/s x Sin 20º. The acceleration downward due to gravity is 9.8 m/s². Take it away!

Answer 4

find the horizontal and vertical components of the velocity, then use kinematic equations.

*Hint* there is no horizontal acceleration.

Answer 5

SHAME ON YOU!!! You should not throw bricks in the air! They may end up hurting someone when they land!!!