Question

A 68 kg bungee jumper is standing on a platform 46 m above the ground. The jumper falls from rest and it is assumed that the bungee cord, which is attached to the bungee jumper at one end and the platform at the other, has no effect for the first 9 m of the fall, i.e. the natural length of the cord is 9 m. When the bungee jumper is more than 9 m away from the platform the cord acts as an ideal spring with a spring constant k=66 N/m. In the following you may assume that the only forces acting on the bungee jumper are his weight and, when the cord is stretched beyond it natural length, the elastic force of the bungee cord. (a) What is his speed after falling 9 m from the platform? (b) What is his speed after falling 31 m from the platform? (c) Sketch the shape of the position-time graph of the bungee jumper.

Answer 1

h0 = 46m

Ha = 9m

Hb = 31m

k = 66 N/m^2 and g =9.80m/s^2 , m= 68 kg

a) Speed after falling 9m

1/2 m(Va^2) + mg(Ha) = mg(h0)

Va = Root (2g (h0-Ha))

= Root [2*9.8m/s^2 (46m-9m)]

Va = 26.9 m/s

b) Speed after falling 31m

mg(h0) =mg(Hb)+k[(Ha)-(Hb)]^2/2 +mVb^2/2

68kg*9.8m/s^2*46m = 68kg * 9.8m/s^2 (31m) + 66N/m[9-31]^2 /2 + 68*Vb^2 /2

30654.4 = 9996+31944+ 34Vb^2

Vb^2 = (30654.4-9996-31944)/34 = -331.92

Vb = 18.21 m/s

c) sketch are attached below