If the net vertical impulse of a 580 n person is 165 ns over 0.5 s, what was the average net vertical force produced? if her initial vertical velocity was –0.200 m/s, what was her vertical take-off velocity?

1) The average force delivered is the impulse divided by the elapsed time:

F = J/Δt

F is the average force, J is the impulse, and Δt is the elapsed time.

Given values:

J = 165Ns

Δt = 0.5s

Plug in the values and solve for F:

F = 165/0.5

F = 330N

2) The impulse J is the person's change in momentum Δp:

J = Δp

The person's momentum is the product of their mass and velocity:

p = mv

Assuming the person's mass stays constant, their change in momentum is equal to their mass multiplied by their change in velocity:

Δp = mΔv

By definition the change in velocity is the difference of the final and initial velocities:

Δv = v_{f} - v_{i}

Knowing all this, we therefore have this equation:

J = m(v_{f} - v_{i})

We do not know the person's mass yet, but we do know her weight, and the weight is given by:

W = mg

W is the weight, m is the mass, and g is the acceleration due to earth's gravity near its surface

Given values:

W = 580N

g = 9.81m/s²

Plug in the values and solve for m:

580 = m×9.81

m = 59.1kg

Now to find the take-off velocity we use this equation:

J = m(v_{f} - v_{i})

Given values:

J = 165Ns

m = 59.1kg

v_{i} = -0.200m/s

Plug in these values and solve for v_{f}:

165 = 59.1(v_{f} + 0.200)

v_{f} + 0.200 = 2.79

v_{f} = 2.59m/s

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explanation:

bc its just b

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