physics max speed and height?

Question

The Royal Gorge bridge over the Arkansas
River is 372 m above the river. A bungee
jumper of mass 110 kg has an elastic cord of
length 79 m attached to her feet. Assume the
cord acts like a spring of force constant k. The
jumper leaps, barely touches the water, and
after numerous ups and downs comes to rest
at a height h above the water.
The acceleration of gravity is 9:81 m=s2 :
Find h. Answer in units of m.
part2
Find the maximum speed of the jumper. An-
swer in units of m=s.

Answer 1

part1(complicated)
to find k use F=kx
110(9.81)=k(372-79)
k=3.68
end energy
PEspring=.5(3.68)s^2
PEgravity=(110)(9.81)(372-79-s)
KE=.5(110)0^2...=0
beginning energy
PEgravity=(110)(9.81)(372-79)
PEspring=.5(3.68)0^2...=0
KE=.5(110)(39.37)^2
energy is conserved, so
(.5(3.68)s^2)+(110)(9.81)(372-79-s)=((110)(9.81)(372-79))+(.5(110)(39.37)^2)
1.84s^2+316176.3-1079.1s=316176.3+85248.9
1.84s^2-1079.1s-85248.9=0
quadratic formula...
s=70.5(stretched)
h=372-(79+70.5)
h=222.5m

part2(easy)
vf^2=vi^2+2as
vf^2=0+2(9.81)(79)
vf=39.37m/s

Answer 2

The horizontal velocity is V*cos41º. and the horizontal distance is x = V*cos41º*tf, where tf is the flight time. Initial vertical velocity is V*sin41º. the time to reach max height hm is tm = V*sin41º/g. The time to reach ground is the same so the flight time is 2*tm = 2*V*sin41º/g. x = 2*(V²/g)*sin41º*cos41º V = √[1.4*g/(2*sin41º*cos41º)] = 3.72 m/s hmax = V*sin41º*t - 0.5*g*tm² V*sin41º/g hmax = 0.5*(V*sin41º)²/g = 0.031 hm = 0.30 m