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Punchlinegirl
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A bungee jumper (m = 75.00 kg) tied to a 48.00 m cord, leaps off a 78.00 m tall bridge. He falls to 8.00 m above the water before the bungee cord pulls him back up. What size impulse is exerted on the bungee jumper while the cord stretches?
I found the velocity by
v_final^2= V_inital + 2ad
V^2= 2(9.8)(30)
V=24.2 m/s
Impulse= change in momentum.
I think the initial momentum is 0 since he isn't moving.
Impulse= m*v
= 1818.7 kg*m/s
This wasn't right.. can someone help me out?
Thanks
I found the velocity by
v_final^2= V_inital + 2ad
V^2= 2(9.8)(30)
V=24.2 m/s
Impulse= change in momentum.
I think the initial momentum is 0 since he isn't moving.
Impulse= m*v
= 1818.7 kg*m/s
This wasn't right.. can someone help me out?
Thanks