average force

(a) Use conservation of energy
(1/2)mv^2 = mgh
where m is tha mass of Hu (=58 kg)
v is the speed just before entering into water
g is the acceleration due to gravity
h is the height of fall(=10m)

(1/2)(58)v^2 = 58x10x10
v^2 = 200 (m/s)^2, thus v = 14.14 m/s

(b) Use (work done) = change of kinetic energy
i.e. (net force)x distance = change of kinetic energy
(F-mg).s = (1/2)mv^2
where s is the distance travelled into water
F is the force exerted by water

(F - 58x10)x4 = (1/2)(58)(14.14^2)
solve for F gives F = 2030 N

a)v^2-u^2=2gs,v=14.14ms-1

b)v^2-u^2=2(a+g)s
0-14.14^2=2(a+10)*4
a=-35ms-2

F=ma
F=58*-35
F=-2030N
所以the magnitude of the average force exerted on him by the water=2030N