A rock perched on the edge of a cliff falls off. It starts from rest, and falls freely for 1.48 seconds before reaching the ground at the base of the cliff. What is the speed of the stone just before it hits the ground?
v=g*time
To find the speed of the stone just before it hits the ground, we can use the equation for freely falling objects:
v = g * t
where:
- v is the final velocity (speed),
- g is the acceleration due to gravity (which is approximately 9.8 m/s² on Earth),
- t is the time the object has been falling.
In this case, the stone falls freely for 1.48 seconds. Plugging in the values, we get:
v = 9.8 m/s² * 1.48 s
Simplifying the equation:
v = 14.504 m/s
Therefore, the speed of the stone just before it hits the ground is approximately 14.5 m/s.