A small rock is thrown vertically upward with a speed of 13.0from the edge of the roof of a 37.0 tall building. The rock doesn't hit the building on its way back down and lands in the street below. Air resistance can be neglected.
A) What is the speed of the rock just before it hits the street?
Total energy at start = (1/2) m Vi^2 + m g h
Total energy at finish = (1/2) m v^2
no losses so
(1/2) m v^2 = (1/2) m (13)^2 + m g (37)
v^2 = 169 + 2(9.81)(37)
To find the speed of the rock just before it hits the street, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (speed just before it hits the street)
u = initial velocity (speed at the start)
a = acceleration
s = displacement
Since the rock is thrown vertically upward, the initial velocity (u) would be positive 13.0 m/s (upward direction). The acceleration (a) due to gravity is always directed downward and is approximately 9.8 m/s^2. The displacement (s) is equal to the height of the building, which is 37.0 m.
Now, let's use the equation of motion to find the final velocity (v):
v^2 = (13.0 m/s)^2 + 2(-9.8 m/s^2)(-37.0 m)
v^2 = 169.0 m^2/s^2 + 722.8 m^2/s^2
v^2 = 891.8 m^2/s^2
Taking the square root of both sides to solve for v:
v = √891.8 m^2/s^2
v ≈ 29.87 m/s
Therefore, the speed of the rock just before it hits the street is approximately 29.87 m/s.