Question

A bungee jumper of mass 80kg jumps from a platform.

Answer 1

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)

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

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

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