A water park has a slide that finishes with the rider flying horizontally of the bottom of the slide. The slide is designed to end 1.5 m above the water level, and the average rider is estimated to leave the bottom of the slide at 25 m/s. How far Horizontally will the rider travel from the bottom of the slide before hitting the water?

25 m/s * √(1.5 m/g) ≈ 9.78 m

Explanation:

When effects of air resistance is neglected, the motion can be modeled as

x(t) =  25 m/s * t

y(t) = 1.5 m - g * t²

based on a coordinate system where the x axis is on the water level (modeled as a plane), y axis being parallel to the gravitational force and the point of origin sitting at the point on the water surface directly below the exit of the slide. The x value is the horizontal travel we’re interested in. Time t begins at the exit of the slide by this rider. g is the gravitational acceleration.

At the time when the rider hits water, we know y = 0.

0 = 1.5 m - g * t² ⇔ t = ±√(1.5 m/g)

Since we are only interested in the non-negative (not past) time range, it’s t = √(1.5 m/g)

The horizontal travel is thus:

x(√(1.5 m/g)) = 25 m/s * √(1.5 m/g) ≈ 25 * √(1.5/9.81) m ≈ 9.78 m

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y