Brock Brown

2022-01-07

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.

Bukvald5z

Step 1
By neglecting the resistance of the air, both stones will experience an acceleration only on the vertical axis which is the acceleration of the gravity, and the velocity in the horizontal axis is constant. Therefore, the following equation describes the motion of the stones in the vertical axis:
$y={v}_{0y}t+\frac{1}{2}a{t}^{2}$
since both stones are thrown horizontally, the vertical initial velocity is zero, so:
$y=\frac{1}{2}a{t}^{2}$
the first stone will make a vertical displacement magnitude equals the height of the first building ${H}_{1}$ but with negative sign, during a time of flight ${t}_{1}$, so:
$-{H}_{1}=\frac{1}{2}a{t}_{1}^{2}$
the second stone will make a verticaldisplacement magnitude equals the height of the second building ${H}_{2}$ but with negative sign, during a time of flight ${t}_{2}$, so:
${H}_{2}=\frac{1}{2}a{t}_{2}^{2}$
Step 2
The distance in the horizontal axis is given by:
$x={v}_{0x}t$
$t=\frac{x}{{v}_{0x}}$ (1)
Both stones are thrown horizontally with the same velocity (say ${v}_{x}$), so for the first stone this equation will become:
${t}_{1}=\frac{{x}_{1}}{{v}_{x}}$
and for the second stone is,
${t}_{2}=\frac{{x}_{2}}{{v}_{x}}$
hence, the height of the two building will become:

but the first stone lands twice as far from the base of the base of the building which means ${x}_{1}=2{x}_{2}$, so:

clearly ${H}_{1}>{H}_{2}$, so the ratio, so the height of the taller building to the height of the shorter building is:
$\frac{{H}_{1}}{{H}_{2}}=\frac{-\frac{1}{2}a{\left(\frac{2{x}_{2}}{{v}_{x}}\right)}^{2}}{-\frac{1}{2}a{\left(\frac{{x}_{2}}{{v}_{x}}\right)}^{2}}$

Papilys3q

Distance from the base of the building to place where stone landing:
$l=vt$
where v is velocity, t is time.
Time can be found from:
$h=\frac{{gt}^{2}}{2}$
where h is height of the building
$t=\sqrt{\frac{2h}{g}}$
Therefore:
${l}_{1}=v\sqrt{\frac{2{h}_{1}}{g}},{l}_{2}=v\sqrt{\frac{2{h}_{1}}{g}}$
$\frac{{l}_{1}}{{l}_{2}}=\sqrt{\frac{{h}_{1}}{{h}_{2}}}=2$
$\frac{{h}_{1}}{{h}_{2}}=4$

karton

4 times taller
Explanation:
The horizontal distance travelled by a projectile is
$d={v}_{x}t\left(1\right)$
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
$y=h-\frac{1}{2}g{t}^{2}$
where h is the initial height, $g=9.8m/{s}^{2}$ 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
$0=h-\frac{1}{2}g{t}^{2}$
$t=\sqrt{\frac{2h}{g}}$
Substituting into (1),
$d={v}_{x}\sqrt{\frac{2h}{g}}$
This can be re-arranged as
${d}^{2}={v}_{x}^{2}\frac{h}{2g}$
$h=\frac{2g{d}^{2}}{{v}_{x}^{2}}$
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 ${h}^{\prime }=\frac{2g\left(2d{\right)}^{2}}{{v}_{x}^{2}}=4\left(\frac{2g{d}^{2}}{{v}_{x}^{2}}\right)=4h$
so the taller building is 4 times taller than the smaller one.

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