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To solve this problem, we can use the kinematic equation for vertical motion:

h = ut + (1/2)gt^2

where: h = height (distance) fallen u = initial velocity g = acceleration due to gravity (approximately -32 feet per second squared) t = time

In this case, the height fallen is 96 feet, the initial velocity is 16 feet per second, and the acceleration due to gravity is -32 feet per second squared. We need to find the time it takes until the diver enters the water, so we can set h = 0 and solve for t.

0 = ut + (1/2)gt^2

0 = 16t + (1/2)(-32)t^2

0 = 16t - 16t^2

Rearranging the equation:

16t^2 - 16t = 0

Factor out 16t:

16t(t - 1) = 0

This equation has two solutions: t = 0 and t = 1. However, t = 0 represents the initial time when the diver is still on the ledge, so we can ignore that solution.

Therefore, the diver will enter the water after 1 second.

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