How much time to evade a falling object? Free-fall word problem

Question

A woman accidentally drops a flowerpot from the rooftop of a 25-meter building. A 1.5 meters tall man walking notices the falling flowerpot above him at about 4 meters above the ground. How much time does the man have to evade it?

I think the formula to be used is y = υ₀t + (1/2)gt² but I'm not sure if it's right.

Answer 1

At 25 meters:

(In the woman’s hands at the top of the building)

height h = 25m

initial velocity vi = 0m/s (not falling yet)

acceleration gravity a = -9.8m/s^2

Let’s use the information at 25m to find out the time for the flower pot to be 4m above the ground:

h initial = 25m

h final = 4m

together these give displacement: hf - hi = -21m

(This will be our delta y) y = -21m

v initial = 0m/s

a = -9.8 m/s^2

We have y, vi, and a. We need t.

Use y = Vi*t + 1/2*at^2 to find t/

-21 = 0*t + 1/2(-9.8)t^2

-21 = 0 - 4.9t^2

4.29 = t^2 square root both sides

t = 2.07 seconds.

Next, we will repeat this process but we will find how long it takes the flower pot to fall to the height of 1.5m: the location of the man’s head.

h initial = 25m

h final = 1.5m

use these to find displacement

hf - hi = y 1.5-25 =y

y = -23.5 m

vi = 0m/s

a = -9.8m/s^2

y = vi *t + 1/2*at^2

-23.5 = 0*t + 1/2(-9.8)t^2

-23.5 = 0 - 4.9t^2

4.80 = t^2 square root both sides

t = 2.19 seconds

Now, we can use this information to find how much time passes between when the man notices the flower pot (when it’s 4m above) and when the flower pot reaches the height of his head (1.5m).

We can imagine the time in segments:

Time from 25m-4m : time from when the pot is dropped to when the man notices it.

time = 2.07 seconds

Time from 4m-1.5m : time that the man has to react before the pot reaches his head.

time = ???

Time from 25m - 1.5m : total time

total time = 2.19 seconds

Total time = time(dropping pot to noticing pot) + time(noticing pot to hitting head)

2.19 = 2.07 + ??? Subtract 2.07 from both sides

time = 0.12 seconds

The man has 0.12 seconds to react to the pot. He will most certainly be hit on the head.

Answer 2

That is the right formula to calculate the distance the flowerpot

falls in time t, provided at time 0 it had the speed of u₀.

In your case y = 2.5 m -- the distance between flower pot and the man's head

at the moment he notices the flower pot.

From that you can calculate how much time it will take for the flower pot

to hit his head; you can calculate it, provided you know u₀.

To calculate u₀ let's use the variable τ to represent the time it took

the flower pot to fall from rest to the height of 4 m, i.e., to fall 21 m.

(We use τ to avoid confusion with t, which is the time between being

at height 4m and hitting his head.)

Using your formula:

21 = (1/2)gτ²

So

τ = √(42/g)

Then we can use the relation

u₀ = gτ = g√(42/g) = √(42g)

Plugging everything into your equation

2.5 = √(42g)t + (1/2)gt²

Rewrite it into more convenient quadratic equation

gt² + √(168g)t - 5 = 0

It has two solutions

t = (-√(168g) ± √(168g + 20g))/2g =

(-√(168) ± √(168 + 20))/2√g =

(-12.96 ± 13.71)/2√g

The negative solution is not physically meaningful.

The positive solution is

(-12.96 + 13.71)/2√g = 0.12 s

Answer 3

We will use the equation y(t)=y0+v0*t+(1/2)*a*t^2 at three critical points. The first is when the woman drops the pot at t=0. From this point, we see that y0=25m, v0=0m/s, and a=-9.8m/s^2. The next point is when the man sees the pot, which is when y(t)=4m. From plugging in at that point, we get 4=25-0*t-4.9*t^2 or 4.9t^2-21=0 and solve for t to get t=2.07s. The last point is when the flower pot is at y=1.5m, which is where it would hit the man. We use the same process as the last point to get 1.5=25-0*t-4.9t^2 or 4.9t^2 -23.5=0, which gives t=2.19s. The man has the amount of time between noticing the pot and when it would hit him to get out of the way, so that would be 2.19-2.07=0.12s