A person drops a small rock from a bridge. If it takes 1.70 seconds for the rock to reach the water below the bridge, how high was the bridge above the water? please explain.
The distance it falls in t seconds after being dropped is
H = (1/2) g t^2
g is the acceleration of gravity, which you must have learned about by now. It is 9.8 m/s^2 on Earth.
Set t = 1.7 seconds and use the formault to solve for the height, H, in meters.
To calculate the height of the bridge above the water, we can use the equations of motion under constant acceleration. Here's how you can find the answer step by step:
1. Define the knowns:
- Time taken (t) = 1.70 seconds
- Acceleration due to gravity (g) = 9.8 m/s² (assuming the rock is dropped near the surface of the Earth)
2. Identify the variables we need to find:
- Initial velocity (u)
- Height of the bridge (h)
3. Recall the equation of motion:
h = ut + (1/2)gt²
4. Plug in the values:
h = (0)t + (1/2)(9.8)(1.70)²
Simplify: h = (1/2)(9.8)(2.89)
5. Calculate:
h = 14.057 meters
So, the height of the bridge above the water is approximately 14.057 meters.