Question

in the above diagram, the long jumper's take-off velocity is 12.2 m/s at an angle of 30o from the horizontal. her center of mass is 1.6 m above the ground at take-off and is 0.15 m above the ground at landing. neglect air resistance, find how far does his center of mass displace horizontally between take-off and landing (in m).

Answer 1

Sy = Vy t - 1/2 g t^2

Vy = 12.2 * sin 30 = 6.1 m/s vertical take-off speed

Vx = 12.2 * cos 30 = 10.6 m/s

H = Vy t - 1/2 g t^2

(.15 - 1.6) = 6.1 t - 4.9 t^2 substituting assuming uniform CM speed

4.9 t^2 - 6.1 t -1.45 = 0

t^2 - 1.24 t - .296 = 0 divide by 4.9

t = 1.44 s using max t

Sx = 10.6 m/s * 1.44 m = 15.3 m

(this is ridiculous since Sx = 50 ft whereas the world record is < 30 ft

Check using range formula with equal start/finish height

R = v^2 sin 60 / 9.80 = 13.2 m something appears wrong with problem

(Note: 12.2 m/s = 40 ft / sec and 300 ft (100yds) / 40 = 7.5 sec for 100 yds)