A rock is held steady over a cliff and dropped. 1.1 seconds later, another rock is thrown straight down at a speed of 12.1 m/s, and hits the first rock. How far have the rocks dropped before they collide? How long is the first rock in the air before it gets hits by the second rock?
To find the distance the rocks have dropped before they collide, we need to calculate the distance each rock has traveled individually.
Let's start with the first rock that was dropped. We can use the equation of motion for free fall:
d = (1/2) * g * t^2
where d is the distance, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Plugging in the values:
d1 = (1/2) * 9.8 * (1.1)^2
d1 ≈ 5.3905 meters
So, the first rock has dropped approximately 5.3905 meters before the second rock is thrown.
Now let's find the time it takes for the second rock to hit the first rock. Since we are only interested in the time for the first rock, let's ignore the time it took for the second rock to be thrown.
To calculate the time it takes for the first rock to get hit, we can use the equation of motion:
d = ut + (1/2) * a * t^2
where d is the distance, u is the initial velocity (0 m/s since it was dropped), a is the acceleration due to gravity (9.8 m/s^2), and t is the time.
Rearranging the equation:
t^2 = (2d) / a
t = sqrt(2d / a)
Plugging in the values:
t = sqrt(2 * 5.3905 / 9.8)
t ≈ 0.7798 seconds
So, the first rock is in the air for approximately 0.7798 seconds before it gets hit by the second rock.