Question

A 72.0 kg swimmer jumps into the old swimming hole from a tree limb that is 3.90 m above the water.

A. Use energy conservation to find his speed just as he hits the water if he just holds his nose and drops in.
b) Use energy conservation to find his speed just he hits the water if he bravely jumps straight up (but just beyond the board!) at 2.90 m/s .
c) Use energy conservation to find his speed just he hits the water if he manages to jump downward at 2.90 m/s .

Answer 1

Step-by-step explanation:

The Law of Energy Conservation states that K1 + U1 = K2 + U2

m= 72.0 kg

h= 3.90 m

a)

K1 + U1 = K2 + U2

0 + mgh = 1/2mvf^2 + 0

mass cancels out so gh=1/2vf^2

(9.8 m/s^2)(3.9 m)=(.5)(vf^2)

vf= 8.74 m/s

b)

1/2mv^2 + mgh = 1/2mv^2 + 0

mass cancels again

(.5)(2.9^2 m/s) + (9.8 m/s^2)(3.9 m) = (.5)(vf^2)

vf= 9.21 m/s

c)

This would be the same as the past problem as the velocity gets squared so direction along the axis doesn't matter. Thus, vf= 9.21 m/s