Question

A woman accidentally drops a flowerpot from a windowsill at a height d above the street towards a man of height h standing below. The woman calls out to the man in just enough time for the man to move out of the way. If the man needs a time interval of Δt to respond to the warning, at what height above the street will the flowerpot be when the woman calls out the warning? (Use the following as necessary: d, h, Δt, v for the speed of sound, and g for gravitational acceleration.)

Answer 1

Answer:

h^2 - ( 2t_o v_s + 2v_s^2 /g) h + v_s^2  \  t_o^2 =0

The correct result is that of a positive height

Explanation:

For this exercise we use the kinematic relations, let's start by finding the time it takes for the sound to reach the man

             v_s = y / t

             t = $\frac{y}{ v_s}$

this height is y = h

             t =  \frac{h}{ v_s}

the man has a response time of t = t₀, therefore

time to move is

             t' =  t - t₀

             

the initial height of flower pot is

           y = y₀ + v₀ t' - ½ g  t'²

when it reaches the floor the height is zero y = 0 and as the pot is dropped its initial velocity is zero v₀ = 0

            0 = y₀ +0 - ½ g (t -t₀)²

if the initial height is i = h,

             

            h = ½ g ($\frac{h}{v_s}$ - t₀)²2

            $\frac{2}{g} h$ = $\frac{h^2}{v_s^2}$ - $\frac{2t_o }{v_s} h$ + t₀²

            $\frac{h^2}{v_s^2} - ( \frac{2t_o}{v_s} + \frac{2}{g} ) h + t_o^2 = 0$h2 / vs2 - (2nd / vs + 2 / g) h + to2 - = 0

            $h^2 - ( 2t_o v_s + 2v_s^2 /g) h + v_s^2 \ t_o^2 =0$

 

To know the height, you must solve the second degree equation, it is much easier with numerical values.

             The correct result is that of a positive height