A falling flower pot (kinematics question), just need slight correction

• Sean1218
In summary, the problem involves a flowerpot falling past a window of height 1.9m in 0.2s. Using the equations d = v1t + 1/2at^2, a = (v2 - v1) / t, and d = (v2 + v1) / 2 * t, the solution involves finding the distance from the starting point to the top of the window (d) and the time taken by the flowerpot to reach the top of the window (t). By setting up two equations (1) and (2), the values for d and t can be solved for, giving a final answer of d = 3.7m and t = 0
Sean1218

Homework Statement

A falling flowerpot takes 0.2s to fall past a window that is 1.9m tall. From what height above the top of the window was the flower pot dropped? g = -9.8 m/s^2, v1 = 0 m/s

Homework Equations

d = v1t + 1/2at^2
a = (v2 - v1) / t
d = (v2 + v1) / 2 * t

The Attempt at a Solution

Ok, so, I sort of got the right answer, but I'm doing something wrong with my negative signs.

I plugged the values into the first equation to solve for v2 (speed at the bottom of the window), then plugged the values into the second equation (this time with v1 as 0 m/s (speed at where flower pot is dropped i.e. rest)), and got a negative value.

(8.52 - 0) / -9.8 = t = - 0.87

Then I plugged that time value into the 3rd equation to solve for d, which would give me a negative height value (-3.7, while the answer is 3.7). So, where did I mess up with my negatives?

If you have taken g negative then d and v must be negative.

Try this approach.

Let d be the distance from the starting point to the top of the window.

Then the distance between the starting point and the bottom of the window is ( d + 1.9 m). If t is the time taken by the flowerpot to reach the top of the window, then

d = 1/2*g*t^2 ...(1)

(d + 1.9 ) = 1/2*g*(t + 0.2 s)^2...(2)

Solve these two equations to find d and t.

1. What is the initial velocity of the flower pot?

The initial velocity of the flower pot is the speed at which it is dropped from its original height.

2. How do you calculate the time it takes for the flower pot to hit the ground?

The time it takes for the flower pot to hit the ground can be calculated using the formula t = √(2h/g), where h is the initial height and g is the acceleration due to gravity.

3. What is the final velocity of the flower pot right before it hits the ground?

The final velocity of the flower pot right before it hits the ground is equal to the initial velocity plus the product of acceleration due to gravity and the time it takes to reach the ground.

4. Does air resistance affect the motion of the falling flower pot?

Yes, air resistance does affect the motion of the falling flower pot. It causes a drag force that acts opposite to the direction of motion and slows down the pot.

5. How does the mass of the flower pot impact its motion?

The mass of the flower pot does not affect its motion in terms of acceleration due to gravity. However, a heavier pot will experience a greater force of air resistance, which will slow it down more than a lighter pot.

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