Water contained by a dam exits a spill-way 685ft above the base of the dam. Determine the velocity of the water as it strikes the base of the dam if its initial velocity as it leaves the spillway is negligible.
v = at
so, how long does it take to fall 685 ft?
To determine the velocity of the water as it strikes the base of the dam, we can use the principle of conservation of energy.
At the top of the spillway, the water has potential energy due to its height above the base of the dam. As it falls, this potential energy is converted into kinetic energy.
The potential energy (PE) of an object is given by the formula: PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height from which the object is falling.
In this case, the mass of the water is not given, but we can ignore it because we only need to find the velocity. So, we can write the equation for potential energy as: PE = gh.
The kinetic energy (KE) of an object is given by the formula: KE = 1/2 mv^2, where v is the velocity of the object.
According to the principle of conservation of energy, the potential energy at the top of the spillway is equal to the kinetic energy at the base of the dam, neglecting any other losses or gains of energy. So we can write the equation: PE = KE.
Substituting the formulas for potential energy and kinetic energy, and canceling out the mass, we get: gh = 1/2 v^2.
Now, we can rearrange the equation to solve for velocity (v): v^2 = 2gh.
Plugging in the values: g = 32.2 ft/s^2 (acceleration due to gravity) and h = 685 ft, we can calculate the velocity.
v^2 = 2 * 32.2 ft/s^2 * 685 ft.
v^2 = 43988 ft^2/s^2.
Taking the square root of both sides, we get:
v = sqrt(43988) ft/s.
Using a calculator, we find that v is approximately 209.8 ft/s.
Therefore, the velocity of the water as it strikes the base of the dam is approximately 209.8 ft/s.