Respuesta :
The integral of the function of the force applied to the time of application
of the force gives the impulse of the force.
The speed with which the ball rebounds from the plate is approximately
34.23 m/s.
Reasons:
Force = Rate of change of momentum
F × t = m·(v₂ - v₁)
Where;
Speed of the ball, v₁ = 16 m/s
v₂ = The speed with which the ball rebounds from the plate
The impulse, I, of the force is given by the area under the curve.
The height of the triangular curve, h = 2,500 N
The base length of the triangular curve = 7 ms.
Area of a triangle = [tex]\frac{1}{2}[/tex] × Base length × Height
Therefore, we have;
[tex]Area \ under \ the \ curve, \, I = \dfrac{0.007 \times 2500}{2} = 8.75[/tex]
8.75 = 0.48 × (v₂ - 16)
v₂ ≈ 34.23 m/s
Therefore, the speed with which the ball rebounds, v₂ ≈ 34.23 m/s.
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