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A bungee jumper of mass 80kg jumps from a platform.
After the rope is fully stretched, the rope provides an
impulse of 3200Ns, propelling him upwards at 10m/s. What
speed was he moving at when he finished falling?

Respuesta :

onyebuchinnaji

The speed he was moving at when he finished falling is 30 m/s.

The given parameters;

mass of the bungee, m = 80 kg

impulse provided by the rope, J = 3200 Ns

initial upward velocity of the jumper, u = 10 m/s

  • Let the final velocity after falling = v
  • Let the upwards motion = negative
  • Let the downwards motion when falling = positive

Apply the principle of conservation of linear momentum;

J = ΔP = Δmv = m(v - u)

3200 = 80(v - (-10))

3200 = 80(v + 10)

[tex]v+ 10 = \frac{3200}{80} \\\\v+ 10 = 40\\\\v = 40-10\\\\v = 30 \ m/s[/tex]

Thus, the speed he was moving at when he finished falling is 30 m/s.

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