Question

Consult Conceptual Example 9 in preparation for this problem. Interactive LearningWare 6.3 also provides useful background. The drawing shows a person who, starting from rest at the top of a cliff, swings down at the end of a rope, releases it, and falls into the water below. There are two paths by which the person can enter the water. Suppose he enters the water at a speed of 15.3 m/s via path 1. How fast is he moving on path 2 when he releases the rope at a height of 2.26 m above the water? Ignore the effects of air resistance.

Answer 1

Path 1

Initial velocity at the top of the cliff is u =0 m/s

let height of the cliff beh

total energy at the top of the cliff ismgh

velocity of person before reaching water is v =15.3 m/s

then total energy at the surface of water= (1/2)mv² therefore,

$\frac{1}{2}mv^{2} = mgh$

$h=\frac{v^{2}}{2g}$

$h=\frac{(15.3m/s)^{2}}{2*9.81m/s^{2}}$

$h=11.93m$

  

Path 2

let height of fall of person when he releasesthe rope is h ' = 2.26 m

let velocity of person at the point of releaseis v '

then total energy is (1/2)mv'² +mgh'

then

$\frac{1}{2}m{v}'^{2} + mg{h}' = mgh$

${v}'^{2}=2g(h-{h}')$

${v}'^{2}=2*9.81m/s^{2}(11.93m-2.26m)$

${v}'^{2}=189.75(m/s)^{2}$

then,

${v}'=13.77m/s$