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