Question

An adult (mass 98.36kg) is standing on top of a 4.45m cliff right next to a river. He grabs a vine of length 30.1m, which the point of support is directly above a senior (mass 57.53kg). The adult grabs the senior at the bottom of his swing and has just enough speed to make it to the cliff on the other side. What would be the adult’s speed right before he catches the senior? What is the maximum height of the cliff that both of them can reach?

Answer 1

Answer:

1. 2.98m/s

2. 0.28m

Explanation:

The energy equation would work great in this scenario:

E=K+U. Since all of our energy comes from gravitational potential energy, and we are interested in finding the kinetic energy, all our mechanical energy must be in kinetic form, therefore:

$mgy=1/2mv^2+0\\98.36*4.45=1/2*98.36*v^2\\437.7=49.18v^2\\8.9=v^2\\2.98=v$

We can use energy to find max height.

For energy, set the equation E=K+U as 437.7=(mass adult+mass senior)gh:

$437.7=155.89gh\\\\2.8=gh\\0.28=h$