Question

A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. Its speed is 26 m/s when it reaches ground level. What was its speed when its height was half that of its starting point?

Answer 1

Answer:

The velocity is  $v_h = 19.2 \ m/s$

Explanation:

From the question we are told that

   The speed of the roller coaster at ground level  is $v = 26 \ m/s$

 

Generally we can define the roller coaster speed at ground level using the an equation of motion as

     $v^2 = u^2 + 2 g s$

 u is  zero given that the roller coaster started from rest

      So

            $26^2 = 0 + 2 * g * s$

So  

           $s = \frac{26^2}{ 2 * g }$

=>       $s = 37.6 \ m$

Now the displacement half way is mathematically represented as

       

    $s_{h} = \frac{37.6}{2}$

     $s_{h} = 18.8 \ m$

So

      $v_h ^2 = u^2 + 2 * g * s_h$

Where  $v_h$ is the velocity at the half way point

=>  $v_h = \sqrt{ 0 + 2 * 9.8 * 18.8 }$

=>   $v_h = 19.2 \ m/s$