A 75 kg man executes a standing jump with the aid of a harness and support wire.

Question

A 75 kg man executes a standing jump with the aid of a harness and support wire. In addition to the constant 1100 N force his legs exert during the push off phase, the wire has a lift mechanism that applies a force given by 550*e^(-s/L) (in Newtons) where L is .4 meters and s is the distance traveled. The lift mechanism is engaged at the bottom of the crouch (where s =0) and a safety catch detaches the wire when the stuntman leaves the ground. A) For crouch depth of .4 meters compute the maximal elevation of his center of gravity. Hint: it may be easiest to work from first principles recalling that a= v dv/dz = 1/2 d(v^2)/dz. B) If the safety catch fails to disengage, show that his center of gravity is elevated .462 meters at the top of the jump. Hint: e a similar approach to part a for the airborne phase.

Explanation / Answer

The time he takes to go up from y(max) /2 to the y{max } is the same as the time he falls from y [max ] to y [ max] /2. Let y max = H and then y max = H/2 t1^2 = 2g H/2. ===========1 Simnilarly, Time it takes him to go from the floor to that height t2^2 = 2g H======================2 1divided by 2 gives (t1/t2)^2 = 1/2 = 0.5 t1/t2 = 0.707 neaarly 70 %. t1 = 70% t2. Thus, for 70% of the total time of going up, he is in the upper half of the total height. Hence he seems to hang in the air

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