Answer
4
Work Step by Step
Let's take,
Height of the taller building = H
Height of the shorter building = h
Horizontal displacement after being thrown from the shorter building = x
Horizontal displacement after being thrown from the taller building = 2x
Initial horizontal velocity = u
Let's apply equation 3.5b $S=ut+\frac{1}{2}at^{2}$ in the vertical direction for both incidents.
Taller building =>
$\downarrow S=ut+\frac{1}{2}at^{2}$ ; Let's plug known values into this equation.
$H=0+\frac{1}{2}(g)t_{1}^{2}=\frac{1}{2}(g)t_{1}^{2}$
Similarly, shorter building =>
$h=\frac{1}{2}(g)t_{2}^{2}$
Let's apply the equation $S=ut$ in the horizontal direction for both incidents.
Taller building =>
$\rightarrow S=ut$ ; Let's plug known values into this equation.
$2x=ut_{1}=> t_{1}=\frac{2x}{u}$
Similarly, shorter building =>
$t_{2}=\frac{x}{u}$
$\frac{H}{h}=\frac{\frac{1}{2}(g)t_{1}^{2}}{\frac{1}{2}(g)t_{2}^{2}}=\frac{t_{1}^{2}}{t_{2}^{2}}$ ; Let's plug known values into this equation.
$\frac{H}{h}=\frac{(\frac{2x}{u})^{2}}{(\frac{x}{u})^{2}}=4$
The ratio of the height of the taller building to the shorter building is 4
