A rock is dropped from rest off a cliff to the surface of a lake which is 6.2 m below. What is the velocity of the rock just as it hits the water?
9.8 m/s ?
using the formula vf^2=vo^2+ 2ad you can find out the answer. Because the stone is dropped the acceleration is -9.8m/s^2 and the vo=0m/s. The distance is aleady given to you. From here simply plug in the numbers and do the calculations to figure out vf. It should be around 11m/s
To find the velocity of the rock just as it hits the water, we can use the equations of motion.
The equation we will use is:
v^2 = u^2 + 2as
Where:
v = final velocity (what we want to find)
u = initial velocity (which is 0 since the rock was dropped from rest)
a = acceleration due to gravity (which is approximately 9.8 m/s^2)
s = displacement (the height from the cliff to the surface of the lake, which is 6.2 m)
Plugging in these values, the equation becomes:
v^2 = 0^2 + 2 * 9.8 * 6.2
Simplifying further:
v^2 = 2 * 9.8 * 6.2
v^2 = 121.36
To find v, we take the square root of both sides:
v = √121.36
v ≈ 11.02 m/s
Therefore, the velocity of the rock just as it hits the water is approximately 11.02 m/s.