Cliff divers at Acapulco jump into the sea
from a cliff 36.4 m high. At the level of the
sea, a rock sticks out a horizontal distance of
13.16 m.
The acceleration of gravity is 9.8 m/s
2
.
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
h = 0.5*g*T^2 = 36.4,
0.5*9.8*T^2 = 36.4,
T = 2.73 s.
Vx*T = 13.16.
Vx*2.73 = 13.16,
Vx =
Why did the cliff divers invite the rock to their performance? Because they wanted to SEA if they could AVOID rock bottom! Now, to answer your question: We can use the kinematic equation to find the minimum horizontal velocity. The equation we'll use is:
d = v_x * t
Where d is the horizontal distance, v_x is the horizontal velocity, and t is the time of flight.
Since we are trying to find the minimum horizontal velocity, we can assume that the cliff divers just barely miss the rock. This means that the time it takes for them to reach the rock should be the same as the time it takes for them to fall the full height of the cliff.
Using the kinematic equation for vertical motion:
y = v_y * t - (1/2) * g * t^2
Where y is the height, v_y is the vertical velocity, g is the acceleration due to gravity, and t is the time of flight.
We have the values:
y = 36.4 m
g = 9.8 m/s^2
Since the divers start with zero vertical velocity, the equation simplifies to:
y = -(1/2) * g * t^2
Solving for t, we get:
t = sqrt((2 * y) / g)
Plugging in the values:
t = sqrt((2 * 36.4) / 9.8)
t ≈ 2.82 s
Now that we have the time of flight, we can find the horizontal distance using the equation:
d = v_x * t
Plugging in the known values:
13.16 m = v_x * 2.82 s
Solving for v_x:
v_x ≈ 13.16 m / 2.82 s
v_x ≈ 4.66 m/s
So, the minimum horizontal velocity the cliff divers must leave the top of the cliff with in order to miss the rock is approximately 4.66 m/s. Keep those divers rockin'!
To find the minimum horizontal velocity, we can use the kinematic equations.
The first step is to determine the time it would take for the divers to fall from the cliff to the sea. We can use the following kinematic equation:
h = (1/2)gt^2
Where:
h = 36.4 m (height of the cliff)
g = 9.8 m/s^2 (acceleration due to gravity)
t = time (unknown)
Rearranging the equation to solve for time:
t = sqrt((2h)/g)
t = sqrt((2 * 36.4) / 9.8)
t = sqrt(3.71)
t ≈ 1.925 seconds
Now, we can use the time calculated to find the minimum horizontal velocity using the formula:
v = d/t
Where:
v = horizontal velocity (unknown, to be determined)
d = 13.16 m (horizontal distance)
v = 13.16 m / 1.925 s
v ≈ 6.82 m/s
Therefore, the minimum horizontal velocity the cliff divers must have to miss the rock is approximately 6.82 m/s.
To determine the minimum horizontal velocity required for the cliff divers to miss the rock, we can use the principles of projectile motion.
First, let's consider the motion of the divers in the vertical direction. The vertical distance the divers need to cover is 36.4 m (the height of the cliff). The only force acting on the divers in the vertical direction is gravity, which causes them to accelerate downward with an acceleration of 9.8 m/s^2. We can use the following equation to calculate the time it takes for the divers to fall from the cliff:
Δy = v_iy * t + 0.5 * a_y * t^2
Where:
Δy = 36.4 m (vertical distance)
v_iy = initial vertical velocity (unknown)
a_y = acceleration due to gravity (-9.8 m/s^2)
t = time of flight
Since the divers start from rest vertically (v_iy = 0), we can simplify the equation to:
Δy = 0.5 * a_y * t^2
Plugging in the values:
36.4 = 0.5 * (-9.8) * t^2
Solving for t^2, we get:
t^2 = -36.4 / (-4.9)
t^2 = 7.4286
t ≈ √7.4286
t ≈ 2.725 s
Now, let's consider the motion of the divers in the horizontal direction. We want the divers to miss the rock, which means they should reach the horizontal distance of 13.16 m. The horizontal distance is directly related to the horizontal velocity and the time of flight:
Δx = v_ix * t
Where:
Δx = 13.16 m (horizontal distance)
v_ix = initial horizontal velocity (unknown)
t = 2.725 s (time of flight)
Rearranging the equation, we can solve for v_ix:
v_ix = Δx / t
v_ix = 13.16 / 2.725
v_ix ≈ 4.828 m/s
Therefore, the minimum horizontal velocity the cliff divers must have when leaving the top of the cliff in order to miss the rock is approximately 4.828 m/s.