Angle Building Bricks

Angle Building Bricks
You have a machine that throws a brick at a speed Vo at an angle above the horizontal Ѳ?

Place the machine near the shore the roof of this building and the angle 30 ° and the initial velocity is 10 m / s. If the brick lands 18m from the base of the building, the height is the building?

You have to think of the initial velocity of the block in terms of its components. Think of X and Y-components: X: 10 cos 30 º = 8.66 m / s Y-component: 10 sin 30 ° = 5 m / s You must first find the amount of time that the brick is in the air by the kinematic equation xf = xi + vi • t + 1 / 2 • A • T ^ 2 where xf is the final position x, xi is the initial position is the initial velocity vi x, a is the acceleration in the x direction, t is time the ball is in the air. We know it is a 0, since we can neglect air resistance. Therefore: xf = xi + vi t • 18 m = 0 m + 8.66 m / s • t = 18 m 8.66 m / stt = 18 • m / 8.66 m / s T = 2.078 seconds Now use the same kinematic equation for the y direction This equation is f = yi + vi • t + 1 / 2 • A • T ^ 2. F is the final position (0 m), yi is what we are looking for (and position), vi is the initial velocity in the direction and this time, a is the acceleration in the y direction (-9.81 m / s ^ 2), and t is the time. We can solve for yi based on what we know. 0 m = Yi + 5 m / s • 2.078 s – 1 / 2 • 9.81 m / s ^ 2 • (2,078 s) ^ 2 = 0 m + i 10.39 m – 4905 m / s ^ 2 • s ^ 4.32 2 0 m = yi + 10.39 m – 21.1896 m 0 m = yi – yi = 10.80 m 10.80 m Thus, the height of the building is 10.80 meters. I hope this explanation helps!

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