Emily holds a banana of mass m over the edge of a bridge of height h. She drops the banana and it falls to the river below. Use conservation of energy to find the speed of the banana just before hitting the water.Express your answer in terms of some or all of the variables h and m, and appropriate constants.

v= the square root of 2gh

2gh

mgh=1/2 m v^2

solve for v

GPE=mgh & KE=1/2m(v^2)

-In this equation potential becomes kinetic-

mgh=1/2m(v^2) [divide both sides by mass so it cancels]
->
mh=1/2(v^2) [divide both sides by 0.5 which is 1/2 so it cancels]
->
2mh=v^2 [find root of both sides so square of v cancels]
->
Square root of (2mh) = velocity

To find the speed of the banana just before hitting the water, we can use the principle of conservation of energy, which states that the total mechanical energy of an object remains constant as long as no external forces are acting on it.

Initially, the banana at the top of the bridge has potential energy due to its height above the water, and no kinetic energy. As it falls, it loses potential energy and gains an equal amount of kinetic energy. Just before hitting the water, all of its initial potential energy has been converted to kinetic energy.

The potential energy of an object is given by the equation PE = mgh, where m is the mass of the banana, g is the acceleration due to gravity, and h is the height of the bridge.

The kinetic energy of an object is given by the equation KE = 0.5mv^2, where v is the speed of the banana just before hitting the water.

Since energy is conserved, we can equate these two expressions:

mgh = 0.5mv^2

Simplifying the equation, we can cancel out the mass 'm' on both sides:

gh = 0.5v^2

Rearranging the equation and solving for v, we get:

v^2 = 2gh

v = sqrt(2gh)

Therefore, the speed of the banana just before hitting the water is given by the square root of 2 times the product of the acceleration due to gravity and the height of the bridge.