Question

You are at the Grand Canyon, standing at the edge of a ledge 1,857 m high. You have a mass of 61 kg. You decide to take a selfie to share with your science teacher when you get home which causes you to wonder... a. How much gravitational potential energy do you have standing at the edge of this cliff?a. How much gravitational potential energy do you have standing at the edge of this cliff? b. If you were to trip and fall mid-selfie, your stored potential energy would be converted to kinetic energy, as you are in motion. Assume that all of the GPE you calculated in part A would be converted to kinetic energy when you fall. If so, how fast would you be falling?

Answer 1

Answer:

A. 1132770 kgm^2sec^-2  B. v = 192.717m/sec.

Explanation:

We Know,

               Potential Enery = mgh

mgh = 10*61*1857[took 10 for ease, can be 9.8]

mgh = 1132770 kgm^2sec^-2.

Now,

        mgh = 1/2mv^2;

         gh = 1/2v^2.

         20*1857 = v^2.

         37410 = v^2.

         so, v = 192.717m/sec.