A stone falls from a cliff. It lands on the ground below the cliff after 4.2 seconds. The height of the cliff can be calculated to be what?
It will be the distance that an object falls in 4.2 s, with zero initial vertical velocity component.
You surely know the formula for that:
Y = (g/2) t^2
t = 4.2 s g = 9.8 m/s^2
To calculate the height of the cliff, you can use the equations of motion. The one that relates distance, time, initial velocity, and acceleration is:
d = vit + (1/2)at^2,
Where:
- d is the distance traveled,
- vi is the initial velocity (which in this case is 0 since the stone is only affected by gravity, not thrown),
- a is the acceleration due to gravity (approximately 9.8 m/s^2),
- t is the time taken.
In this case, you know that the stone takes 4.2 seconds to reach the ground, and assuming the height of the cliff is h, the distance traveled is h. Therefore, you can rewrite the equation as:
h = (1/2)gt^2,
where g is the acceleration due to gravity.
Now, plug in the known values:
h = (1/2)(9.8 m/s^2)(4.2 s)^2.
Calculating this expression will give you the height of the cliff.