A gargoyle statue falls off the side of a building from a height of 75 m. (Assume no air resistance.) How fast is the statue moving when it reaches the ground?
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To determine the speed of the gargoyle statue when it reaches the ground, we use the principles of free fall under the influence of gravity. Since we are assuming no air resistance, the only force acting on the gargoyle is gravity, accelerating it downward.
The formula to calculate the final velocity (\( v \)) of an object in free fall is:
\[ v = \sqrt{2gh} \]
where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, m/s^2 \) on Earth's surface) and \( h \) is the height from which the object falls. Substituting in the given values:
\[ v = \sqrt{2 \times 9.81 \, m/s^2 \times 75 \, m} \]
\[ v = \sqrt{2 \times 9.81 \times 75} \]
\[ v = \sqrt{1471.5} \]
\[ v \approx 38.36 \, m/s \]
So, the gargoyle statue would be moving approximately \( 38.36 \, m/s \) when it reaches the ground.
Extra: Free fall is a particular case of motion where an object is subject only to the force of gravity, neglecting other forces like air resistance. The acceleration due to gravity (\( g \)) is constant near the Earth's surface and has a value of approximately \( 9.81 \, m/s^2 \).
When an object is in free fall, it accelerates downwards at this rate, which means that for every second the object is falling, its velocity increases by \( 9.81 \, m/s \). This constant acceleration allows us to use the kinematic equations of motion to predict the final velocity, total time of fall, and distance fallen over time.
The formula used to calculate the final velocity is derived from the work-energy principle or from kinematic equations. The work-energy principle relates the kinetic energy of the object to the work done by gravity in pulling it down. The kinetic energy of the object when it reaches the ground is equal to the potential energy it had at the height from which it fell. The kinematic equation used is based on the relationship that the final velocity squared is equal to the initial velocity squared plus two times the acceleration times the distance (\( v^2 = u^2 + 2gh \)), where initial velocity (\( u \)) is zero for an object starting from rest.