Final answer:
The speed of the rock just before it hits the street can be calculated using the kinematic equation vf^2 = vi^2 + 2ad, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and d is the distance traveled. Plugging in the given values, the speed is determined to be 26.8 m/s.
Step-by-step explanation:
When a rock is thrown vertically upward, it experiences the force of gravity pulling it back down. As the rock reaches its highest point, its velocity becomes zero before it starts falling back down due to gravity. The speed of the rock just before it hits the street can be calculated using the kinematic equation:
vf^2 = vi^2 + 2ad
where vf is the final velocity, vi is the initial velocity, a is the acceleration (which is equal to -9.8 m/s^2 for objects falling near the surface of the Earth), and d is the distance traveled.
Since the rock is thrown vertically upward, the distance traveled is equal to the height of the building, which is 51.5 m. We can plug in the given values:
vf^2 = (26.8 m/s)^2 + 2(-9.8 m/s^2)(51.5 m)
Simplifying this equation gives us: vf^2 = 719.04 m^2/s^2
Taking the square root of both sides gives us the final velocity: vf = 26.8 m/s
So, the speed of the rock just before it hits the street is 26.8 m/s.