To solve for the magnitude of the average force that the water exerts on the diver, let's break the problem into steps:

1. Calculate the Diver's Speed Just Before Hitting the Water

   • We’ll use energy conservation from the top of the tower to the water surface, since only gravity acts during the fall.

   $mgh = \frac{1}{2}mv^2$

   where:

   • $m = 65.0 \, \text{kg}$
   • $g = 9.81 \, \text{m/s}^2$
   • $h = 4.10 \, \text{m}$
   • $v$ is the speed just before hitting the water.

   Solving for $v$:

   $v = \sqrt{2gh} = \sqrt{2 \cdot 9.81 \cdot 4.10} \approx 8.97 \, \text{m/s}$

2. Calculate the Deceleration Underwater

   • Now that we know her speed at the water's surface, we can calculate the deceleration as she comes to a stop 1.30 m underwater.

   We’ll use the kinematic equation:

   $v^2 = u^2 + 2ad$

   where:

   • $v = 0 \, \text{m/s}$ (final speed underwater, as she comes to rest),
   • $u = 8.97 \, \text{m/s}$,
   • $d = 1.30 \, \text{m}$,
   • $a$ is the deceleration due to the water.

   Rearranging for $a$:

   $a = \frac{v^2 - u^2}{2d} = \frac{0 - (8.97)^2}{2 \cdot 1.30} \approx -30.97 \, \text{m/s}^2$

3. Calculate the Average Force Exerted by the Water

   • Using Newton’s second law, we can now find the average force exerted by the water, $F$, which is equal to $ma$:

   $F = ma = 65.0 \cdot 30.97 \approx 2013 \, \text{N}$

   Thus, the magnitude of the average force exerted by the water on the diver is approximately 2013 N.

Would you like further details on any part of this solution, or do you have any questions?

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Here are some related questions to explore further:

1. How would the force change if the diver’s mass were different?
2. What would be the effect if the water's resistance varied with depth?
3. How would air resistance alter the diver's speed before hitting the water?
4. How would the result differ if the water depth were shallower?
5. What are examples of non-conservative forces in other sports activities?

Tip: In problems involving vertical descent and impact, conserving energy before impact and then applying forces post-impact simplifies the calculation.