A heavy book is launched horizontally out a window from the first floor, a height, h, above the ground, with initial velocity, v0, and it hits the ground a horizontal distance X1 away from the window. Another book is similarly launched (same initial velocity) from the second floor window, a height 2h above the ground. Where does the second book land relative to the first book

x₂ / x₁ = √2

Explanation:

To solve this exercise we can use the projectile launch ratios, let's find the time it takes for the second book to reach the ground

             y = y₀ + t - ½ g t²

as the book is thrown horizontally v_{oy} = 0, when it reaches the ground its height is zero y= 0

            0 = y₀ - ½ g t²

            t =

            t = \sqrt{    \frac{2 \ 2h}{ g} }

with this time we calculate the horizontal distance traveled

            x = v₀ t

            x₂ = v₀

now let's calculate the time it takes him to get to the floor when he leaves from the first floor

           t =\sqrt{    \frac{2y_o}{ g} }

the horizontal distance traveled is

           x₁ = v₀

therefore the difference in distance between the two runs is

           Δx = x₂-x₁

           Δx = v₀ \sqrt{ \frac{4h}{g} } - v₀ \sqrt{ \frac{2h}{g} }

            Δx = v₀ \sqrt{ \frac{2h}{g} }    √2

            Δx =√2    x₁

the relationship between the two distances is

             x₂ / x₁ = √2

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.