Question

You are lying on a beach, your eyes 20cm above the sand. Just as the Sun sets, fully disappearing over the horizon, you immediately jump up, your eyes now 150 cm above the sand, and you can again just see the top of the Sun. If you count the number of seconds until the Sun fully disappears again, you can estimate the Earth’s radius. But for this Problem, use the known radius of the Earth to calculate the time t.

Answer 1

Answer:

The  time is  $t = 8.78 \ s$

Explanation:

From the question we are told that  

  The  initial height of the eye  is  $h _1 = 20 \ cm = 0.2 \ m$

  The  height of the eye when you jumped up is $h_2 = 150 \ cm = 1.5 \ m$

 An illustration of this question is shown on the first uploaded image

Generally the radius of the earth is  $R = 6.38*10^{6} \ m$

Now from the diagram first sun means first time you saw the sun and  the second sun means second time you saw the sun then

   H is the height increase when you quickly stood up which is mathematically evaluated as

        $H = h_2 -h_1$

        $H = 1.5 - 0.2$

        $H = 1.3 \ m$

Also $\theta$ i the angular displacement between the first and second position and from geometry it is also the angle at  one of the sides of the right angle triangle

Applying Pythagoras theorem

     $(R+H)^2 = K^2 + R^2$

=>   $R^2 + H^2 + 2RH = K^2 + R^2$

Now given that H is very small compared to R the we ignore $H^2$

So

     $R^2 + 2RH = K^2 + R^2$

=>    $K = \sqrt{2RH}$

=>     $K = \sqrt{2 * 6.38*10^6 * 1.3 }$

=>      $K = 4073 \ m$

Now the $\theta$ is mathematically evaluated using SOHCAHTOA as follows

      $tan \theta = \frac{K}{R}$

      $\theta = tan^{-1}[ \frac{K}{R}]$

=>   $\theta = tan^{-1}[ \frac{ 4073}{6.38*10^{6}}]$

=>   $\theta = 0.0366^o$

Generally

  $1 \revolution\ around \ the\ earth = 24 \ hours = 86400 \ seconds = 360 ^o$

So  

    $\frac{\theta}{360} = \frac{t}{86400}$

=>  $\frac{0.0366}{360} = \frac{t}{86400}$

=>  $t = 8.78 \ s$