A diver running 1.6 m/s dives out horizontally from the edge of a vertical cliff and 3.5 s later reaches the water below.How high was the cliff?How far from its base did the diver hit the water?
I answered this last night, check replies.
The cliff is 60m tall
The diver landed 5.6m from it's base
First use velocity formula to find distance of base, then use -(1/2)(-9.80)(3.5^2) to find height of cliff.
To find the height of the cliff and the horizontal distance from the base where the diver hits the water, we will need to use the equations of motion.
Let's break down the problem:
Given:
- Initial velocity of the diver, u = 1.6 m/s (horizontal velocity)
- Time taken to reach the water, t = 3.5 s
We need to find:
- Height of the cliff, h
- Horizontal distance from the base where the diver hits the water, x
We can start by calculating the vertical displacement of the diver using the formula:
Displacement (vertical) = u * t + 0.5 * a * t^2
Since the diver is diving out horizontally, there is no vertical acceleration (a = 0). Thus, the equation simplifies to:
Displacement (vertical) = u * t
Plugging in the values:
Displacement (vertical) = 1.6 m/s * 3.5 s = 5.6 m
This means the height of the cliff is 5.6 meters.
Next, we can calculate the horizontal distance covered by the diver. The horizontal distance can be found using the formula:
Distance (horizontal) = u * t
Plugging in the values:
Distance (horizontal) = 1.6 m/s * 3.5 s = 5.6 m
Therefore, the diver hits the water 5.6 meters from the base of the cliff.
In summary:
- The height of the cliff is 5.6 meters.
- The diver hits the water 5.6 meters from the base of the cliff.