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

the answer is the first electromagnet (a). this is because it has more coils around the ferromagnetic material. increasing the number of coils increases the surface area in which the metal interacts with the current in the coil. this enables more atoms in the metal to become aligned hence becoming more magnetized.