Question

A person jumps out a fourth-story window 14 m above a firefighter safety net. The survivor stretches the net 1.8 m before coming to rest. what was the deceleration experienced by the survivor? Use g = 9.8 m/s2 Calculate to one decimal.

Answer 1

Answer:

The deceleration is  $a = - 76.27 m/s^2$

Explanation:

From the question we are told that

   The height above  firefighter safety net is $H = 14 \ m$

   The length by which the net is stretched is $s = 1.8 \ m$

   

From the law of energy conservation

    $KE_T + PE_T = KE_B + PE_B$

 Where $KE_T$ is the kinetic energy of the person before jumping which equal to zero(because to kinetic energy at maximum height )

   and  $PE_T$ is the potential energy of the before jumping  which is mathematically represented at

          $PE_T = mg H$

and  $KE_B$ is the kinetic energy of the person just before landing on the safety net  which is mathematically represented at

        $KE_B = \frac{1}{2} m v^2$

and  $PE_B$ is the potential energy of the person as he lands on the safety net which has a value of zero (because it is converted to kinetic energy )

   So the above equation becomes

          $mgH = \frac{1}{2} m v^2$

=>           $v = \sqrt{2 gH }$

    substituting values

                $v = 16.57 m/s$

Applying the equation o motion

             $v_f = v + 2 a s$

Now the final velocity is zero because the person comes to rest

      So

         $0 = 16.57 + 2 * a * 1.8$

            $a = - \frac{16.57^2 }{2 * 1.8}$

            $a = - 76.27 m/s^2$