A high jumper, falling at -4m/s lands on a foam pit and comes to rest, compressing the pit .40m. If the pit is able to exert an avg force of 1200N on the high jumper in breaking the fall, what is the athletes mass???
2006-11-07 20:16:38 · 3 answers · asked by Neo Sapien 1 in Science & Mathematics ➔ Physics
since v^2=v0^2+2a(x-x0)
0^2=16+ 2a(.4)
a=20m/s^2
F=ma
1200=m(20)
m=60kg
2006-11-07 20:33:17 · answer #1 · answered by igot4onit 2 · 0⤊ 0⤋
Find the deceleration. Initial speed is Vo = 4 m/s and final speed is V= 0 so:
V^2 = Vo^2 - 2*a*x, 0 = 4^2 - 2*a*0.4, a = 20 m/s
ΣF = F - w, where F = 1200N and w the weight, w = m*g (g: the acceleration of gravity, g = 10 m/s^2 approx)
ΣF = m*a, so
m*a = F - m*g, m(a+g) = F, m = F/(a + g), m = 1200/(20 + 10),
m = 40 kg
2006-11-08 06:00:48 · answer #2 · answered by Dimos F 4 · 0⤊ 0⤋
around 100 kg
2006-11-08 04:24:02 · answer #3 · answered by Tyza H 2 · 0⤊ 0⤋