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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

Respuesta :

Answer:

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 + [tex]v_{oy}[/tex] 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 = [tex]\sqrt{ \frac{2y_o}{ g} }[/tex]

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

with this time we calculate the horizontal distance traveled

            x = v₀ t

            x₂ = v₀ [tex]\sqrt{ \frac{4h}{g} }[/tex]

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 [tex]\sqrt{ \frac{2h}{g} }[/tex]

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

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