Divers for many years have dived into the ocean from the cliffs of Acapulco, a distance of 35 m above the water. How fast are the divers moving when they hit the water? If the divers travel a distance of 5 m through the water before momentarily coming to rest, what acceleration do the divers experience while in the water?
Thanks!
To find the speed at which the divers hit the water, we can use the principles of motion. We'll assume there is no air resistance.
1. Start by identifying the given information:
- Distance fallen from the cliff, h = 35 m
- Distance traveled through the water, d = 5 m
- Acceleration due to gravity, g = 9.8 m/s^2
2. Calculate the speed at which the divers hit the water:
We can use the equation for free fall:
v^2 = u^2 + 2as
Here, u represents the initial velocity (which is 0 since the divers start from rest) and s represents the distance fallen.
v^2 = 0 + 2gh
v^2 = 2gh
v = √(2gh)
Substituting the given values:
v = √(2 * 9.8 * 35) = √(686) ≈ 26.2 m/s
Therefore, the divers are moving at a speed of approximately 26.2 m/s when they hit the water.
3. Calculate the acceleration experienced by the divers while in the water:
We can use the equation of motion:
v^2 = u^2 + 2as
Rearranging it to solve for a:
a = (v^2 - u^2) / (2s)
Here, v is the final velocity, u is the initial velocity, and s is the distance traveled.
a = (0 - v^2) / (2d)
a = -v^2 / (2d)
Substituting the given values:
a = - (26.2^2) / (2 * 5)
a = -682.44 / 10
a ≈ -68.244 m/s^2
The acceleration experienced by the divers while in the water is approximately -68.244 m/s^2, indicating deceleration.
Note: The negative sign indicates that the acceleration is in the opposite direction to the velocity of the divers.
To find out the speed of the divers when they hit the water, we can use the principles of motion under constant acceleration.
Let's assume that the initial height of the divers is h = 35 m, and the final distance they travel in the water is d = 5 m.
To determine the speed of the divers when they hit the water, we need to find their final velocity. We can use the equation of motion:
vf^2 = vi^2 + 2ad
Where:
vf is the final velocity
vi is the initial velocity (which is 0 in this case because the divers start from rest)
a is the acceleration experienced by the divers
d is the distance traveled
Rearranging the equation, we get:
vf = sqrt(2ad)
Substituting the known values into the equation:
vf = sqrt(2 * 9.8 m/s^2 * 5 m)
vf = sqrt(98 m^2/s^2)
vf ≈ 9.90 m/s
Therefore, the divers' speed when they hit the water is approximately 9.90 m/s.
Now, let's calculate the acceleration experienced by the divers while in the water.
We can use the equation:
vf^2 = vi^2 + 2ad
Since the divers come to rest in the water before reaching the final distance d, their final velocity vf is 0. The initial velocity vi is the velocity the divers had when they hit the water, which we calculated to be approximately 9.90 m/s.
Substituting the known values into the equation:
0 = (9.90 m/s)^2 + 2a * 5 m
Rearranging the equation, we get:
2a * 5 m = -(9.90 m/s)^2
a ≈ -(9.90 m/s)^2 / (2 * 5 m)
a ≈ -48.01 m/s^2
Therefore, the acceleration experienced by the divers while in the water is approximately -48.01 m/s^2. The negative sign indicates a deceleration or slowing down.
Note: The negative sign in the acceleration arises from the fact that we have taken the initial direction of motion as positive, and the divers are slowing down in the opposite direction of their initial motion.