What is the speed of a 70kg rock after it has fallen freely for 1000m
v^2 = 2gs
Now plug in your numbers
d = 1/2 * a * t^2 ... t = √(2 d / a) = √(2 * 1000 / 9.8) ... seconds
the speed (velocity) of the rock is time multiplied by gravitational acceleration
To find the speed of the rock after it has fallen freely for 1000 meters, we can use the equations of motion and the laws of physics.
First, let's determine the acceleration due to gravity. On Earth, the acceleration due to gravity is approximately 9.8 meters per second squared (m/s^2). We'll denote this value as "g."
Next, we can use the equation of motion for a freely falling object to find the final velocity (v) of the rock after it has fallen a distance (x) under constant acceleration:
v^2 = u^2 + 2ax
Here:
v is the final velocity (which we're trying to find),
u is the initial velocity (which is zero since the rock starts from rest),
a is the acceleration due to gravity (g),
x is the distance fallen (1000 m).
Substituting these values into the equation, we get:
v^2 = 0 + 2 * g * x
v^2 = 2 * g * x
Now, let's calculate the final velocity (v):
v^2 = 2 * 9.8 m/s^2 * 1000 m
v^2 = 19600 m^2/s^2
To get the value of v, we need to take the square root of both sides:
v = √(19600 m^2/s^2)
v ≈ 140 m/s
Therefore, the speed of the 70kg rock after it has fallen freely for 1000m is approximately 140 meters per second (m/s).