Using conservation of energy, how would you find the speed of a dropped object just before it hits the water?

To find the speed of a dropped object just before it hits the water using conservation of energy, you can equate the potential energy at the height of the drop to the kinetic energy just before impact.

When the object is at height ( h ), its potential energy (PE) can be expressed as ( PE = mgh ), where ( m ) is the mass of the object, ( g ) is the acceleration due to gravity, and ( h ) is the height. As it falls and reaches the water, this potential energy is completely converted into kinetic energy (KE), which can be expressed as ( KE = \frac{1}{2}mv^2 ), where ( v ) is the speed of the object.

Setting the potential energy equal to the kinetic energy gives us:

[ mgh = \frac{1}{2}mv^2 ]

Since ( m ) (the mass of the object) appears on both sides of the equation, it can be canceled out:

[ gh = \frac{1}{2}v^2 ]

To isolate ( v^2 ), you multiply both sides of the equation by 2:

[ 2gh = v^2 ]