Other information: takeoff velocity = (4.7335i +3.686j) m/s horizontal jump dist

Question

Other information:

takeoff velocity = (4.7335i +3.686j) m/s

horizontal jump distance = 3.5607 m

at rest take-off at rest F Jump phase (A): Projectile phase (B) Landing phase (C) Average Impulse due to Non-impulse due to BW 3. When landing, the same jumper decelerates her centre of mass for 0.67s from touch- down to rest (tc). Assuming the centre of mass is at the same height at touch-down and take-off, calculate her a) uniform horizontal and vertical acceleration (acx and acy b) average horizontal and vertical force applied to the ground during landing (FCk Cy

Explanation / Answer

COM is at same height during touch down and takeoff, hence

tc =ta =0.67s

takeof vel = 4.7335 i + 3.686 j m/s

initial horizontal vel = 0

final horiz. vel. = 4.7335 m/s

time t = 0.67 s

v = at

horizontal acceleration acx = 4.7335/0.6 = 7.889 m/s/s

ver. accel. acy = 3.686/0.6 = 6.14 m/s

b) Let be the mass of the jumper, then

Fcx = macx = 7.889 m

Fcy = macy = 6.14 m        

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