A Flowerpot Falling from a Penthouse Suite: Calculating Time to Hit the Ground

The Scenario:

A flowerpot falls off the balcony of a penthouse suite that is 85 meters above the street. The question arises: how long does it take for the flowerpot to hit the ground?

Final Answer:

To calculate the time taken for a flowerpot to hit the ground, we can use the equation of motion and plug in the given values. It takes approximately 4.42 seconds for the flowerpot to hit the ground.

Explanation:

To answer this question, we can use one of the basic equations of motion:
$$d=ut + \\frac{1}{2}at^2$$

Where:
d represents the distance traveled (85 m in this case)
u represents the initial velocity (0 m/s in this case, as the flowerpot is initially at rest)
a represents the acceleration due to gravity (approximately -9.8 m/s^2)
t represents the time taken to hit the ground (which we need to calculate)

Plugging in the values, we have:
$$85 = 0 \\times t + \\frac{1}{2}(-9.8)t^2$$

After simplifying the equation, we can solve for t using the quadratic formula or by factoring:
$$t^2 - \\frac{2(85)}{-9.8} = 0$$
$$t \\approx 4.42 \\text{ seconds}$$

Therefore, it takes approximately 4.42 seconds for the flowerpot to hit the ground.

A flowerpot falls off the balcony of a penthouse suite 85 m above the street. How long does it take to hit the ground? To calculate the time taken for a flowerpot to hit the ground, we can use the equation of motion and plug in the given values. It takes approximately 4.42 seconds for the flowerpot to hit the ground.
← Which object has the greater mass Circular motion understanding angular and tangential speed →