Bang bang, you shot me down...

Classical Mechanics Level 5

A man sitting in a hot air balloon floating h h meters above the ground drops an object towards the ground, and at the same time fires a gunshot.

An observer on the ground, standing on the ground right next to the place of the impact of the object, measures a time difference Δ t = 3 s \Delta t = 3s between the arrival of the sound of the shot and the impact of the object.

What is the sum of the two possible heights h 1 h_{1} and h 2 h_{2} , if both heights are rounded down to the nearest lower integer?

Details and assumptions \textbf{Details and assumptions}

  • The temperature of the air is T = 14.5 ° C T = 14.5°C
  • The ideal gas constant is R = 8.31 J m o l ⋅ K R = 8.31 \frac {J}{mol \cdot K}
  • The molecular mass of air is M M = 28.96 g m o l MM = 28.96 \frac {g}{mol}
  • The adiabatic index of air is 1.4 1.4
  • The gravitational acceleration is g = 9.8 m s 2 g = 9.8 \frac {m}{s^2}
  • For simplicity, round off the speed of sound v s v_{s} to the nearest integer
  • Both heights h 1 h_{1} and h 2 h_{2} are rounded down to the nearest lower integer before being added together
  • Assume no air drag acts upon the object
Image credit: Wikipedia Tommaso.gavioli


The answer is 21551.

Petru Lupsac by Petru Lupsac , 24, Italy

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Petru Lupsac
Dec 3, 2014

First of all, we need to find the speed of sound: v s = 1.4 ⋅ 8.31 J m o l ⋅ K ⋅ ( 273.15 + 14.5 ) K 0.02896 k g m o l ≃ 340 m s v_{s} = \sqrt{\frac{1.4 \cdot 8.31 \frac {J}{mol \cdot K} \cdot (273.15 + 14.5) K}{0.02896 \frac {kg}{mol}}} \simeq \boxed{340 \frac {m}{s}} Note that we had to convert the temperature from ° C °C to K K and the M M MM of the air from g m o l \frac {g}{mol} to k g m o l \frac {kg}{mol} in order to get the correct result.

The time t t required by the sound of the shot to reach the ground observer can be written as t = h v s t = \frac {h}{v_{s}}

Now we can set up the equation to find the two heights: h = 1 2 g ( t + Δ t ) 2 h = \frac 12g(t + \Delta t)^2 Plugging in t t and Δ t \Delta t we get: h = 1 2 g ( h v s + 3 s ) 2 h = \frac12g(\frac hv_{s} + 3s)^2 h = 1 2 g ( h + v s ⋅ 3 s v s ) 2 h = \frac12g(\frac {h + v_{s} \cdot 3s}{v_{s}})^2 h = g 2 v s 2 ( h + v s ⋅ 3 s ) 2 h = \frac {g}{2v_{s}^2}(h + v_{s} \cdot 3s)^2 Rearranging and solving for h h we get: h = v s 2 g − v s ⋅ 3 s ± ( v s 2 g − v s ⋅ 3 s ) 2 − 4 ⋅ 9 v s 2 \large h = \frac {v_{s}^2}{g} - v_{s}\cdot3s \pm \sqrt{(\frac {v_{s}^2}{g} - v_{s}\cdot3s)^2 - 4\cdot9v_{s}^2} Plugging in the values for g g and v s v_{s} and rounding down the two results we get h 1 = 48 m h_{1} = 48 m and h 2 = 21503 m h_{2} = 21503 m .

Therefore, h 1 + h 2 = 48 m + 21503 m = 21551 m h_{1} + h_{2} = 48 m + 21503 m = \boxed{21551 m}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Is value of g=9.8 at that height ?? , what about air drag, how to get a confirmation that the sound reaches the observer without any refraction

Manit Kapoor - 6 years, 6 months ago

Log in to reply

It's more of a kinematics problem, so it is assumed that g is constant (as specified) everywhere, and there is no air drag. Also, I haven't included any information about the object's size or mass, so there is no way you can include air drag. Still, I guess I should add that, thanks!

The sound reaches the observer without refraction because he's standing right next to the place of the impact, which is right underneath the balloon, so even if refraction were considered, it wouldn't have a significant effect.

Petru Lupsac - 6 years, 6 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...