Falls freely from rest and strikes the ground. during the collision with the ground, he comes to rest in a time of 0.040 s. the average force exerted on him by the ground is 18 000 n, where the upward direction is taken to be the positive direction. from what height did the student fall

 y = 2,645 10⁴ / m²

m=80 kg,  y = 4.13 m

Explanation:

We must solve this problem in two parts, one when it is in free fall and another for the collision with the floor

Let's start by analyzing the crash with the floor,

Initial instant When it arrives but if you start to stop

              p₀ = m v

Final moment. When he stopped

              = 0

The momentum is related to the moment by

            I = Δp = p_{f} –p₀

            F t = 0 - mv

            v = -F t / m

Let's calculate

             v = -18000 0.040 / m

             v = -720 / m

The sign indicates that the speed goes down

Now we use energy conservation at two points

Lowest point. Just before crashing

               Em₀ = K = ½ m v²

Highest point. From where it began to fall

               Em_{f} = U = m g y

  Energy is conserved in the fall

               Em₀ = Em_{f}

                ½ m v² = m g y

                y = ½ v² / g

               y = ½ (720 / m)² /9.8

               y = 2,645 10⁴ / m²

For an explicit height value, the object's mass must be known, suppose the masses are m = 80 kg

              y = 2,645 10⁴/80²

              y = 4.13 m

i think it because equal matrices have equal dimensions, only square matrices can be symmetric. for all indices and . every square diagonal matrix is symmetric, since all off-diagonal elements are zero. in linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space.

0.19 m/s and 0.45 m/s