3

I am having difficulty proving this physics formula. It states:

If a physics student drops a watermelon from the roof of a building and hears the watermelon going "splat" on the ground after time $\Delta t$, show that the height of the building $h$ measured in meters is given by

$$\frac{1}{2}g\Bigg(\frac{-{v_s}+\sqrt{{v_s}^2+2g{v_s}{\Delta t}}}{g}\Bigg)^2$$

where $v_s$ is the speed of sound, g is the acceleration due to gravity, and air resistance is ignored.

My attempt went as follows:

Since the initial velocity of the watermelon is zero and $\Delta t$ is equal to the duration of the fall $+$ the time taken for sound to travel the distance $h$, it follows that

$${\Delta x}={v_0}t-\frac{1}{2}gt^2$$

becomes

$$h=\frac{1}{2}g\Big({\Delta t}-\frac{h}{v_s}\Big)^2$$

where $h$ is positive.

Solving for h gives

$$h=\frac{1}{2}g\Big({\Delta t}^2-\frac{2h{\Delta t}}{v_s}+\frac{h^2}{{v_s}^2}\Big)$$

$$h=\frac{1}{2}g{\Delta t}^2-\frac{gh{\Delta t}}{v_s}+\frac{gh^2}{{2v_s}^2}$$

$$2{v_s}^2h={v_s}^2g{\Delta t}^2-2{v_s}gh{\Delta t}+gh^2$$ $$gh^2-2{v_s}(g{\Delta t}+{v_s})h+{v_s}^2g{\Delta t}^2=0$$

$$h=\frac{2{v_s}(g{\Delta t}+{v_s})\pm\sqrt{(2{v_s}(g{\Delta t}+{v_s}))^2-4g{v_s}^2g{\Delta t}^2}}{2g}$$

which clearly doesn't look like the answer above.

Any help?

Anson Pang
  • 589

1 Answers1

3

Your approach is correct but it is not the simplest. You need one additional step: to prove that your result is equivalent to the result that you are trying to prove. And this could easily be the most complicated part of your solution! If you need Mathematica to do it, something is definitely wrong.

This is a classical example how a different approach can greatly simplify the problem.

Take a closer look at the result that you are trying to prove. It looks like:

$$h=\frac g2 (something)^2\tag{1}$$

What is that "something"? The watermelon takes time $\Delta t_1$ to reach the ground:

$$h=\frac g2 (\Delta t_1)^2\tag{2}$$

Compare (1) and (2) and you see that that "something" is actually the time it takes watermelon to reach the ground. So let's try not to solve the problem directly for $h$. Let us solve it for $\Delta t_1$ first. After that, just replace the solution into (2) and you are done!

Sound coming back to the student gives us the second equation:

$$h=v_s\Delta t_2$$

$$\Delta t_2=\frac{h}{v_s}=\frac{g(\Delta t_1)^2}{2v_s}$$

The total time is:

$$\Delta t = \Delta t_1+\Delta t_2$$

$$\Delta t = \Delta t_1+\frac{g(\Delta t_1)^2}{2v_s}$$

It leads to the following (quite simple) quadratic equation:

$$g(\Delta t_1)^2+2v_s\Delta t_1-2v_s\Delta t=0$$

...with the following solutions:

$$\Delta t_1=\frac{-2v_s\pm \sqrt{4v_s^2+8gv_s\Delta t} }{2g}$$

...or if you drop meaningless negative solution:

$$\Delta t_1 =\frac{-v_s+\sqrt{v_s^2+2gv_s\Delta t} }{g}\tag{3}$$

Now replace (3) into (2) and you get:

$$h=\frac{g}{2}\left( \frac{-v_s+\sqrt{v_s^2+2gv_s\Delta t} }{g}\right)^2$$

...which is exactly the result that you wanted to prove. Much simpler, isn't it?

Saša
  • 15,906