A daring 510-N swimmer dives off a cliff with a running horizontal leap. What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is 1.75 m wide and 9.00 m below the top of the cliff?
I do not care if she is thin or fat.
falls 9 meters, how long?
h= (1/2) g t^2
18 = 9.8 t^2
t = 1.36 falling
goes 1.75 meters horizontal in 1.36 seconds
1.75 = u t = u (1.36)
u = 1.29 m/s horizontal speed
To determine the minimum speed the swimmer must have, we need to analyze the motion of the swimmer and calculate the required horizontal velocity.
First, let's consider the vertical motion. The swimmer starts at the top of the cliff and falls downward, reaching a height of 9.00 m below the top of the cliff. We can use the kinematic equation for vertical motion:
Δy = Vyi * t + (1/2) * g * t^2
Where:
Δy is the vertical displacement (9.00 m)
Vyi is the initial vertical velocity (0 m/s, as the swimmer dives perpendicular to the ground)
g is the acceleration due to gravity (-9.8 m/s², assuming downward direction)
t is the time of flight
Since we want to determine the initial horizontal velocity, we only need to consider the vertical displacement. Let's rearrange the equation to solve for t:
9.00 m = 0 * t + (1/2) * (-9.8 m/s²) * t^2
9.00 m = -4.9 m/s² * t^2
t^2 = (9.00 m) / (-4.9 m/s²)
t^2 ≈ -1.84 s²
This means that the time of flight would be imaginary or non-existent, which is not possible. So, we conclude that the swimmer cannot make it to the ledge by just diving off the cliff.
However, assuming that the question is asking for the minimum horizontal speed required to reach the water well beyond the ledge, we can calculate it using the horizontal distance and the time it takes for the swimmer to fall vertically.
The horizontal distance traveled can be estimated by multiplying the time it takes for the swimmer to fall (t) by the required horizontal velocity (Vx):
Δx = Vx * t
Given that the horizontal distance is 1.75 m, we can substitute this value into the equation:
1.75 m = Vx * t
Now, we need to solve for t using the kinematic equations:
t = Δy / Vy
Since the swimmer falls straight down, Vy is equal to the final vertical velocity, which is the square root of 2 times the product of the acceleration due to gravity and the height:
Vy = √(2 * g * Δy)
Vy = √(2 * -9.8 m/s² * 9.00 m)
Vy ≈ 13.43 m/s
Now we can substitute the values of Δy and Vy into the equation for t:
t = 9.00 m / 13.43 m/s
t ≈ 0.67 s
Substituting the obtained value of t back into the equation for horizontal distance:
1.75 m = Vx * 0.67 s
Now we can solve for Vx:
Vx = 1.75 m / 0.67 s
Vx ≈ 2.61 m/s
Therefore, the minimum speed the swimmer must have just as she leaves the top of the cliff is approximately 2.61 m/s horizontally to avoid hitting the ledge.
To determine the minimum speed the swimmer must have as she leaves the top of the cliff, we can use the principle of projectile motion.
Step 1: Identify the known values
- Gravitational acceleration (g) = 9.8 m/s^2 (constant)
- Height difference (h) = 9.00 m
- Horizontal distance (d) = 1.75 m
- Vertical acceleration (ay) = -g (negative because it opposes the positive direction)
- Unknown value: initial horizontal velocity (vx0)
Step 2: Break down the motion into vertical and horizontal components
Since the horizontal motion is independent of the vertical motion, we can analyze them separately.
- Vertical motion: The swimmer starts from rest at the top of the cliff and falls downward, accelerating due to gravity.
- Horizontal motion: The swimmer has a constant horizontal velocity throughout her jump.
Step 3: Determine the time of flight (t)
Using the vertical motion equation:
h = (1/2) * ay * t^2
Substituting the known values:
9.00 m = (1/2) * (-9.8 m/s^2) * t^2
Solving for t:
t^2 = (2 * 9.00 m) / (9.8 m/s^2)
t^2 = 1.8367 s^2
t ≈ 1.36 s
Step 4: Calculate the required horizontal velocity (vx)
Using the horizontal motion equation:
d = vx * t
Substituting the known values:
1.75 m = vx * 1.36 s
Solving for vx:
vx ≈ 1.29 m/s
Therefore, the swimmer must have a minimum horizontal speed of approximately 1.29 m/s as she leaves the top of the cliff in order to miss the ledge at the bottom.