Cliff divers at Acapulco jump into the sea
from a cliff 30.1 m high. At the level of the
sea, a rock sticks out a horizontal distance of
9.19 m.
The acceleration of gravity is 9.8 m/s
With what minimum horizontal velocity
must the cliff divers leave the top of the cliff if
they are to miss the rock?
Answer in units of m/s.
time of flight ... t = √(2 h / g)
min horiz vel = 9.19 m / t
To find the minimum horizontal velocity that cliff divers must leave the top of the cliff with in order to miss the rock, we can use the equations of motion.
First, let's calculate the time it takes for the divers to reach the horizontal distance of 9.19 m. We can use the equation:
distance = initial velocity × time + (1/2) × acceleration × time^2
Since the initial velocity is 0 (since the divers start from rest horizontally), and the acceleration is also 0 in the horizontal direction, the equation simplifies to:
9.19 m = 0 × t + (1/2) × 0 × t^2
Simplifying further, we get:
9.19 m = 0
This equation shows that it takes 0 seconds for the divers to reach the horizontal distance of 9.19 m. This means that they should start with a horizontal velocity that will allow them to reach this distance instantly.
Next, let's determine the time it takes for the divers to fall to the sea level. We can use the equation:
height = (1/2) × acceleration × time^2
Plugging in the values, we get:
30.1 m = (1/2) × 9.8 m/s^2 × t^2
Simplifying, we get:
30.1 m = 4.9 m/s^2 × t^2
Now, let's solve for t:
t^2 = (30.1 m) / (4.9 m/s^2)
t^2 = 6.1429 s^2
t ≈ √6.1429 s
t ≈ 2.48 s
So, it takes approximately 2.48 seconds for the divers to fall to the sea level.
Finally, we can find the minimum horizontal velocity using the equation:
velocity = distance / time
Plugging in the values, we get:
velocity = 9.19 m / 2.48 s
velocity ≈ 3.705 m/s
Therefore, the minimum horizontal velocity the cliff divers must leave the top of the cliff with is approximately 3.705 m/s in order to miss the rock.