a diver running at 1,8 m.s dives out horizontally from the edge of a high vertical cliff. He hits the water 2 seconds later. How high was the cliff?
so it took 2 seconds to fall.
hf=hi-9.8/2 * t^2 hf=0, solve for hi.
YOu really ought to memorize these equations. Hundreds of millions before you have done so.
first you want to
To determine the height of the cliff, we can use the equation of motion for the vertical direction.
The given information includes:
- Initial vertical velocity (u) = 0 m/s (since the diver runs horizontally)
- Time taken to hit the water (t) = 2 s
- Acceleration due to gravity (g) = 9.8 m/s^2 (assuming Earth's gravity)
We can use the equation: s = ut + (1/2)gt^2, where s is the vertical distance covered.
Substituting the values, we get:
s = 0 * 2 + (1/2) * 9.8 * (2^2)
s = 0 + (1/2) * 9.8 * 4
s = 0 + 4.9 * 4
s = 19.6 m
So, the height of the cliff is 19.6 meters.