Calculating Time for a Flowerpot to Fall from a Penthouse Suite

How long does it take for a flowerpot to hit the ground when falling from a penthouse suite 85 m above the street? Final answer: A flowerpot falls from a penthouse suite 85m above street level. Assuming no air resistance, by using the equation of motion h=(1/2)gt², we find that it takes approximately 4.2 seconds for the flowerpot to hit the ground.

When analyzing a scenario where a flowerpot falls from a penthouse suite 85 meters above the street, we encounter a classic problem in physics - free fall under gravity. In this situation, we assume that there is no air resistance affecting the fall of the flowerpot.

The formula used to calculate the time it takes for an object to fall under gravity is h = (1/2)gt², where h represents the height or distance fallen, g is the acceleration due to gravity (approximately 9.8 m/s² on Earth), and t is the time taken for the object to fall.

Given that the flowerpot falls from a height of 85 meters, we can substitute h = 85m and g = 9.8 m/s² into the formula h = (1/2)gt². By solving for t, we find that it takes approximately 4.2 seconds for the flowerpot to hit the ground.

This calculation does not take into account factors such as air resistance, which could affect the time taken for the flowerpot to fall. However, in a simplified scenario of free fall under gravity, the time calculated provides an estimate of how long it would take for the flowerpot to reach the ground.

Understanding the principles of free fall and gravity helps in solving such problems accurately. By applying the equation of motion and considering the given parameters, we can determine the time it takes for objects to fall under the influence of gravity.

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