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

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s

you get 2.3=1.21*\frac{a}2 or 4.6=1.21a or a=3.802 (b)

explanation: