A rock is dropped off of a bridge into the water below. If the initial velocity was zero, and the well is on Earth (g = 9.8 m/s2), and you see the splash 4.59 seconds after the rock was dropped, how high was the bridge above the water? Give your answer in meters, to the nearest tenth of a meter.
h = (1/2)(9.8)(4.59)^2
To find the height of the bridge above the water, we need to use the basic principles of motion and the equation of motion under gravity:
h = (1/2) * g * t^2
where:
- h is the height of the bridge above the water
- g is the acceleration due to gravity (9.8 m/s^2 on Earth)
- t is the time the rock takes to hit the water (4.59 seconds)
Plugging in the given values, we can calculate the height of the bridge:
h = (1/2) * 9.8 * (4.59)^2
h = 1/2 * 9.8 * 21.0281
h = 102.04029 m
Rounding this to the nearest tenth of a meter:
h ≈ 102.0 meters
Therefore, the height of the bridge above the water is approximately 102.0 meters.