A diver running 1.3 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 2.5 s later. How high was the cliff and how far from its base did the diver hit the water
how far can an object fall in 2.5sec
h=1/2 g t^2
What distance does the diver travel horizontally in that time? d=vh*t
30.7
To find the height of the cliff and the horizontal distance from its base where the diver hits the water, we can use the equations of motion.
Let's denote:
- Initial vertical velocity as u = 0 (since the diver starts from rest)
- Acceleration due to gravity as a = 9.8 m/s² (assuming no air resistance)
First, let's find the vertical distance the diver falls. We can use the equation:
s = ut + (1/2)at²
Where s is the vertical distance, u is the initial velocity, t is the time, and a is the acceleration due to gravity.
Plugging in the values we have:
s = 0(2.5) + (1/2)(9.8)(2.5)²
s = (1/2)(9.8)(6.25)
s = 30.625 m
Therefore, the height of the cliff is 30.625 meters.
Next, let's find the horizontal distance the diver travels. Since the diver is moving horizontally with a constant velocity, we can calculate the distance using the formula:
d = v × t
Where d is the horizontal distance, v is the horizontal velocity, and t is the time taken.
Plugging in the values we have:
d = 1.3 × 2.5
d = 3.25 m
Therefore, the diver hits the water 3.25 meters from the base of the cliff.