A daring 510 { N} swimmer dives off a cliff with a running horizontal leap, as shown in the figure below.

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?

Calculate the time to fall 9.00 meters, and call it T

T = sqrt (2H/g)= 1.355 s

Require that
V* T > 1.75 m

Vmin occurs when V = (1.75 m)/T
= 1.29 m/s

mely, check your 1-26,12:52am post.

To calculate the minimum speed required for the swimmer to miss the ledge, we need to consider the horizontal distance she will cover during her horizontal leap and the time it takes for her to fall vertically.

Let's start by calculating the time it takes for the swimmer to fall vertically from the top of the cliff to the height of the ledge.

Using the kinematic equation for vertical motion:
v_f^2 = v_i^2 + 2as

Where:
v_f = final velocity (0 m/s at the height of the ledge)
v_i = initial velocity (0 m/s at the top of the cliff)
a = acceleration due to gravity (-9.8 m/s^2, since we are considering downward motion)
s = displacement (9.00 m, the height of the ledge below the top of the cliff)

Rearranging the equation, we have:
0 = 0 + 2(-9.8)(9)

Simplifying, we find:
0 = -176.4

Since we're solving for time, we know that time is 0 when the swimmer starts falling, and the final time is when the swimmer reaches the ledge. Therefore, we ignore the negative solution.

Now, let's determine the horizontal distance covered during the swimmer's leap.

Using the formula:
d = v * t

Where:
d = horizontal distance covered (1.75 m, the width of the ledge)
v = horizontal velocity of the swimmer
t = time of flight

Rearranging the equation, we have:
v = d / t

Substituting the values we have:
v = 1.75 / 0

Since we cannot divide by zero, it means the swimmer needs to leave the cliff with an infinite horizontal speed to miss the ledge.

Therefore, the minimum speed required for the swimmer as she leaves the top of the cliff is an infinite horizontal speed.

To determine the minimum speed the swimmer must have just as she leaves the top of the cliff, we can use the principles of projectile motion.

First, let's break down the motion into horizontal and vertical components:

1. Horizontal motion: The horizontal component of the swimmer's initial velocity will remain constant throughout the motion. The swimmer must clear the 1.75 m wide ledge, so the horizontal displacement can be taken as 1.75 m.

2. Vertical motion: The vertical component of the swimmer's initial velocity will change due to the acceleration due to gravity. The swimmer must clear the 9.00 m height difference between the top of the cliff and the bottom ledge.

Now, let's analyze the vertical motion separately:

The vertical motion can be described using the following equation of motion:

y = y0 + v0y * t + (1/2) * a * t^2

Where:
y = final height (0 m at the bottom)
y0 = initial height (9.00 m at the top)
v0y = initial vertical velocity (what we need to find)
a = acceleration due to gravity (approximately -9.8 m/s^2, negative because it acts downward)
t = time of flight (unknown)

Since we are looking for the minimum initial velocity, we can assume the swimmer's motion is just barely reaching the edge of the cliff. At this point, the swimmer's final height, y, will be zero. Therefore, we can rewrite the equation as:

0 = 9.00 m + v0y * t + (1/2) * (-9.8 m/s^2) * t^2

Next, we can use the kinematic equation for vertical displacement during free fall:

y = y0 + v0y * t + (1/2) * a * t^2

Since the swimmer ends up at the same height as where she started, the vertical displacement (y - y0) is zero. The equation becomes:

0 = v0y * t + (1/2) * (-9.8 m/s^2) * t^2

Now let's consider the horizontal motion:

The horizontal component of the initial velocity, v0x, will remain constant throughout the motion. We know the horizontal displacement, x, is 1.75 m.

The horizontal displacement x can be calculated using the formula:

x = v0x * t

where v0x is the horizontal component of the initial velocity and t is the time of flight.

Since the horizontal component does not accelerate, its velocity (v0x) remains constant. Therefore, we can rewrite the equation as:

x = v0x * t
1.75 m = v0x * t

From here, we can solve for t:

t = (1.75 m) / v0x

Now we can substitute this value of t into the vertical motion equation:

0 = v0y * [(1.75 m) / v0x] + (1/2) * (-9.8 m/s^2) * [(1.75 m) / v0x]^2

Simplifying further:

0 = (v0y / v0x) * (1.75 m) - (4.9 m/s^2) * [(1.75 m) / v0x]^2

Now, we can rearrange the equation to isolate v0y:

(v0y / v0x) * (1.75 m) = (4.9 m/s^2) * [(1.75 m) / v0x]^2

v0y = (4.9 m/s^2) * [(1.75 m) / v0x]^2 / (1.75 m)

Finally, we can substitute the known values for the horizontal component v0x and solve for v0y:

v0y = (4.9 m/s^2) * [(1.75 m) / v0x]^2 / (1.75 m)

Given that the horizontal speed v0x is not provided, we would need more information to calculate the minimum speed required for the swimmer to clear the ledge.