A 0.4 kg rock is projected from the edge of the top of a building with an initial velocity of 8.16 m/s at an angle 52◦ above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 11.8 m from the base of the building.
How tall is the building? (in m) Assume the ground is level and that the side of the building is vertical. The acceleration of gravity is 9.8 m/s^2
To find the height of the building, we can use the equations of motion for projectile motion.
First, we need to find the time it takes for the rock to reach the ground. We can use the horizontal distance and the initial horizontal velocity for this calculation.
The horizontal velocity (Vx) remains constant throughout the projectile motion. It is given by Vx = initial velocity * cos(angle).
So, Vx = 8.16 m/s * cos(52°) = 5.261 m/s.
Now, using the equation for horizontal distance, we have:
Horizontal distance (dx) = Vx * time.
Given dx = 11.8 m, we can rearrange the equation to solve for time (t).
t = dx / Vx = 11.8 m / 5.261 m/s = 2.24 s (rounded to 2 decimal places).
Now that we have the time it takes for the rock to reach the ground, we can use the equation for vertical distance to find the height of the building.
Vertical distance (dy) = initial vertical velocity * time + (0.5 * acceleration due to gravity * time^2).
The initial vertical velocity (Vy) is given by Vy = initial velocity * sin(angle).
So, Vy = 8.16 m/s * sin(52°) = 6.443 m/s.
Now, substituting the values into the equation:
dy = 6.443 m/s * 2.24 s + (0.5 * 9.8 m/s^2 * (2.24 s)^2).
dy = 6.443 m/s * 2.24 s + (0.5 * 9.8 m/s^2 * 5.0176 s^2).
dy = 14.45 m + 49.159 m.
dy = 63.609 m.
Therefore, the height of the building is approximately 63.61 meters.