You drop a rock from rest out of a window on the top floor of a building, 50.0 m above the ground. When the rock has fallen 3.80 m, your friend throws a second rock straight down from the same window. You notice that both rocks reach the ground at the exact same time. What was the initial velocity of the rock that your friend threw? I honestly just have no idea how to find that answer.
To find the initial velocity of the rock that your friend threw, you can use the kinematic equation for motion under constant acceleration. This equation relates the final velocity (Vf), initial velocity (Vi), acceleration (a), and displacement (d).
The equation is:
Vf^2 = Vi^2 + 2ad
In this case, we can assume that the displacement (d) for both rocks is the same because they both hit the ground. Let's call this displacement "y" for simplicity.
For the first rock, Vf1 = 0 (since it reaches the ground from rest) and d1 = 50.0 m. For the second rock, Vf2 = 0 (since it also reaches the ground) and d2 = 50.0 - 3.80 = 46.2 m (since it falls 3.80 m less than the first rock).
Now, we can set up and solve two equations based on this information:
Equation 1: 0 = Vi1^2 + 2a * 50.0 (for the first rock)
Equation 2: 0 = Vi2^2 + 2a * 46.2 (for the second rock)
Since a is the same for both rocks (gravity), we can see that the only unknown in these equations is Vi2, the initial velocity of the second rock.
To solve these equations simultaneously, we can subtract Equation 1 from Equation 2:
0 = Vi2^2 + 2a * 46.2 - (Vi1^2 + 2a * 50.0)
0 = Vi2^2 - Vi1^2 + 2a (46.2 - 50.0)
Since Vi1 = 0 (for the first rock), the equation simplifies to:
0 = Vi2^2 + 2a (46.2 - 50.0)
Substituting the value of acceleration due to gravity (approximately 9.8 m/s^2) into the equation:
0 = Vi2^2 + 2 * 9.8 (46.2 - 50.0)
Now, we can solve this equation to find the initial velocity of the rock your friend threw (Vi2).