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A water park is designing a new water slide that finishes with the rider flying horizontally off the bottom of the slide. The slide is designed to end 1.2 m above the water level, and the average rider is estimated to leave the bottom of the slide at 25 m/s. How far will the rider fly through the air before hitting the water?

User Arby
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Answer: The rider will fly through the air for about 31.89 meters before hitting the water.

Step-by-step explanation:

To solve this problem, we can use the kinematic equation that relates an object's initial velocity, final velocity, acceleration, displacement, and time, given that the other variables are known.

Here's the equation:

v_f^2 = v_i^2 + 2ad

where

v_f = final velocity = 0 m/s (because the rider stops at the water level)

v_i = initial velocity = 25 m/s

a = acceleration = g = 9.8 m/s^2 (because the only force acting on the rider is gravity)

d = displacement = distance traveled by the rider through the air

We want to solve for d, so we can rearrange the equation to isolate d:

d = (v_f^2 - v_i^2) / 2a

Substituting the values we know, we get:

d = (0^2 - 25^2) / (2 x 9.8)

= -625 / 19.6

= -31.89 m

The negative sign indicates that the displacement is in the opposite direction of the initial velocity, which makes sense since the rider is moving horizontally off the bottom of the slide. To get the absolute value of the displacement, we can take the magnitude:

|d| = 31.89 m

Therefore, the rider will fly through the air for about 31.89 meters before hitting the water.

User Paflow
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