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physics wiz · Apr 2, 2013, 05:37 AM

A bungee jumper falls for 1.3 sec. before the bungee cord began to stretch. Until the jumper has bounced back up to is level, the bungee causes the jumper to have an average acceleration upward of 4m/s^2.
a) How fast is the jumper going when the bungee cord begins to stretch?
b) How far below the diving platform is the jumper at that moment?
c) How long after the bungee cord begins to stretch does the jumper reach the low point of the drop?
d) How far below the diving platform is the jumper at the instant the speed is zero?

Nehushtan · Apr 3, 2013, 07:57 PM

These questions can be solved by using the equations of motion v = u + at and s = ut + ½at².

(a) u = 0 m/s, a = 9.8 m/s², t = 1.3 s, so v = 0 + 9.8×1.3 = 12.74 m/s.

(b) s = 0×1.3 + 0.5×9.8×(1.3)² = 8.281 m.

For (c) and (d), the acceleration is now a = −4 m/s² (the minus sign indicating that it’s upwards).

(c) v = 0 m/s, u = 12.74 m/s (from part (a)), a = −4 m/s², so 0 = 12.74 + (−4)×t giving t = 12.74/4 = 3.185 s.

(d) The distance travelled during the stretching of the cord is s = 12.74×3.185 + 0.5×(−4)×(3.185)² = 20.28845 m, so the distance below the diving platform is found by adding this to the result found in part (b): 20.28845 + 8.281 = 28.56945 m.