Question

Stones are thrown horizontally with the same velocity from the tops of two different buildings. One stone lands twice as far from the base of the building from which it was thrown as does the other stone. Find the ratio of the height of the taller building to the height of the shorter building.
htaller/hsmall=?

Answer 1

Answer:

4 times taller

Explanation:

The horizontal distance travelled by a projectile is

$d=v_x t$ (1)

where $v_x$ is the horizontal velocity (which is constant) and t is the total time of the fall.

The total time of the fall can be found by analyzing the vertical motion of the projectile: the vertical position at time t is given by

$y= h - \frac{1}{2}gt^2$

where h is the initial height, g = 9.8 m/s^2 is the acceleration due to gravity, t is the time.

The total time of the fall is the time t at which y=0 (the projectile reaches the ground), so

$0=h-\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}$

Substituting into (1),

$d= v_x \sqrt{\frac{2h}{g}}$

This can be re-arranged as

$d^2 = v_x ^2 \frac{h}{2g}\\h = \frac{2gd^2}{v_x^2}$

so we see that the initial height depends on the square of the horizontal distance travelled, d.

In this problem, one stands lands twice as far as the other stone does, so

d' = 2d

this means that the height of the taller builing is

$h' = \frac{2g(2d)^2}{v_x^2}=4(\frac{2gd^2}{v_x^2})=4 h$

so the taller building is 4 times taller than the smaller one.