Two rocks are thrown or dropped from a bridge and fall to the water below. The f

Question

Two rocks are thrown or dropped from a bridge and fall to the water below. The first rock has a mass of 1.5 slugs and is thrown upward with an initial velocity of 5 ft/s. The second rock has a mass of 0.75 slugs and is thrown straight down from the bridge with an initial velocity of 1.5 ft/s. Using the concepts of PE and KE and conservation of energy, determine how high above the bridge the first rock goes?

Referring to the situation above and assuming that heights above the bridge are positive and heights below the bridge are negitive : Where, That is , at what position relative to the bridge does Rock #1 have less kinetic energy?

Explanation / Answer

So first we will assume the bridge to have height 0, meaning the potential energy initially is 0. We also know that the rock temporarily stops moving when it reaches its max height, meaning the kinetic energy is 0 there.

KE=PE
(1/2)mv2=mgh

(1/2)v2=gh

h=v2/(2g)=52/(2*32.15)

h=.388 feet

Rock #1 will have less kinetic energy above the bridge (positive height) because as it falls the potential energy is converted into kinetic energy.

Hope that helps

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