Question

A child and sled with a combined mass of 58.0 kg slide down a frictionless slope. If the sled starts from rest and has a speed of 3.20 m/s at the bottom, what is the height of the hill

Answer 1

Answer:

The height is  $h = 0.5224 \ m$

Explanation:

From the question we are told that

   The combined mass of the child and the sled is  $m = 58.0 \ kg$

    The  speed of the sled is  $u = 3.20 \ m/s$

Generally applying SOHCAHTOA on the slope which the combined mass is down from

   Here the length of the slope(L)  where the combined mass slides through  is the hypotenuses

   while the height(h) of the height of the slope is the opposite

Hence from SOHCAHTOA

      $sin (\theta) = \frac{h}{L}$

=>   $Lsin(\theta) = h$

Generally from the kinematic equation we have that

   $v^2 = u^2 + 2aL$

Here the u  is the initial velocity of the combined mass which is zero since it started from rest

 and  a is the acceleration of the combined mass which is mathematically evaluated as

       $a = g * sin (\theta )$

      $v^2 = u^2 + 2 * g * sin (\theta )L$

=>   $2Lsin(\theta ) = \frac{v^2 - u^2 }{g}$

=>   $h = \frac{ v^2 - u^2}{2g}$

=>   $h = \frac{ 3.20^2 - 0^2}{2 * 9.8 }$

=>   $h = 0.5224 \ m$