Math Problem Statement

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?

Solution

To solve this problem, we will break it down into two components: horizontal and vertical motion, treating the swimmer's leap as a projectile motion problem.

Given:

  • Weight of the swimmer W=510NW = 510 \, \text{N}
  • Width of the ledge (horizontal distance to clear) dx=1.75md_x = 1.75 \, \text{m}
  • Height difference between the top and bottom of the cliff (vertical distance) dy=9.00md_y = 9.00 \, \text{m}

The minimum speed required is the swimmer's horizontal velocity as she leaves the cliff. The key concept here is that the horizontal motion is uniform (constant speed), while the vertical motion is accelerated due to gravity.

Step 1: Time to fall

We first calculate the time tt it takes for the swimmer to fall the vertical distance dyd_y of 9.00 meters under the influence of gravity.

Using the kinematic equation for vertical motion: dy=12gt2d_y = \frac{1}{2} g t^2 where:

  • dy=9.00md_y = 9.00 \, \text{m} (vertical distance)
  • g=9.8m/s2g = 9.8 \, \text{m/s}^2 (acceleration due to gravity)

Solving for tt: t=2dyg=2×9.009.8=1.83671.355secondst = \sqrt{\frac{2 d_y}{g}} = \sqrt{\frac{2 \times 9.00}{9.8}} = \sqrt{1.8367} \approx 1.355 \, \text{seconds}

Step 2: Minimum horizontal speed

Now, we calculate the horizontal speed required to cover the horizontal distance dxd_x of 1.75 meters in the time tt found above.

Since horizontal velocity vxv_x is constant, we use the equation: dx=vxtd_x = v_x t Solving for vxv_x: vx=dxt=1.75m1.355s1.29m/sv_x = \frac{d_x}{t} = \frac{1.75 \, \text{m}}{1.355 \, \text{s}} \approx 1.29 \, \text{m/s}

Final Answer:

The swimmer must have a minimum horizontal speed of approximately 1.29 m/s to safely clear the ledge.

Would you like further details or have any questions? Here are five related questions to explore this further:

  1. How would the required horizontal speed change if the height of the cliff increased?
  2. What happens if the swimmer has a higher horizontal speed than the minimum?
  3. How does air resistance affect the motion of the swimmer?
  4. Can we determine the initial velocity of the swimmer if she jumps at an angle?
  5. How would the horizontal speed requirement change if the ledge were wider?

Tip: Always separate the horizontal and vertical components when solving projectile motion problems; they are independent of each other!

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Math Problem Analysis

Mathematical Concepts

Projectile Motion
Kinematics

Formulas

d_y = (1/2) g t^2
d_x = v_x t
v_x = d_x / t

Theorems

Independence of Horizontal and Vertical Motion in Projectile Motion

Suitable Grade Level

Grades 9-12