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A rock is thrown from the roof of a building, with an initial velocity of 10.0 m/s at an angle of 30.0° above the horizontal. The rock is observed to strike the ground 43.0 m from the base of the building. What is the height of the building assuming no air drag?

User Evgenyt
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1 Answer

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Final Answer:

The height of the building is approximately 24.4 meters.

Step-by-step explanation:

Identify the relevant equations:

The horizontal and vertical motions of the rock can be described by the following independent equations:

Horizontal Motion:

x = v_x * t

where:

x is the horizontal distance traveled

v_x is the initial horizontal velocity (10.0 m/s * cos(30°))

t is the time of flight

Vertical Motion:

y = v_y * t - 0.5 * g * t^2

where:

y is the vertical displacement (height of the building)

v_y is the initial vertical velocity (10.0 m/s * sin(30°))

g is the acceleration due to gravity (9.81 m/s²)

Solve for the time of flight:

We are given the horizontal distance traveled (x = 43.0 m). Since we know v_x, we can solve for the time of flight using the horizontal motion equation:

43.0 m = (10.0 m/s * cos(30°)) * t

t ≈ 5.09 s

Solve for the height of the building:

We are given the time of flight (t) and the initial vertical velocity (v_y). We can solve for the height of the building (y) using the vertical motion equation:

y = (10.0 m/s * sin(30°)) * 5.09 s - 0.5 * 9.81 m/s² * (5.09 s)^2

y ≈ 24.4 m

Therefore, the height of the building is approximately 24.4 meters.

User Belek
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