Final answer:
To find the horizontal distance the rock will land from the base of the building, we can use the initial speed and angle of projection. By breaking the initial velocity into horizontal and vertical components, we can determine the horizontal distance traveled by the rock. Plugging in the values and solving the equations, the rock will land approximately 172.8 ft from the base of the building.
Step-by-step explanation:
To solve this problem, we can break the initial velocity of the rock into its horizontal and vertical components. The horizontal component remains the same throughout the motion and is given by the equation vhorizontal = v * cos(θ), where θ is the angle of 60° and v is the initial speed of 80 ft/s. So, vhorizontal = 80 ft/s * cos(60°) = 40 ft/s.
Now, we can use the formula for the vertical displacement of an object in free fall to find the time of flight. The equation is ∆y = vvertical * t + (1/2) * a * t^2, where Δy is the vertical displacement, vvertical is the vertical component of the initial velocity, t is the time of flight, and a is the acceleration due to gravity.
Since the rock is thrown downward, the vertical component of the initial velocity is vvertical = v * sin(θ). Plugging in the values, we have vvertical = 80 ft/s * sin(60°) = 69.3 ft/s.
Using the value of Δy as -168 ft (negative because the displacement is downward), we can rearrange the equation to solve for t: -168 ft = 69.3 ft/s * t + (1/2) * (-32.2 ft/s^2) * t^2. Solving this quadratic equation gives us two possible values for t: t = 0 s and t = 4.32 s.
Since we want to find the horizontal distance traveled by the rock, we can use the formula d = vhorizontal * t. Plugging in the values, we have d = 40 ft/s * 4.32 s = 172.8 ft. Therefore, the rock will land approximately 172.8 ft from the base of the building.