How Does Sound Travel Time Affect Calculating the Fall of a Watermelon?

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Homework Statement


a physics student allows a water melon to fall from a building ( initial velocity = 0) . The student listens the watermelon hitting the floor 2.50 seconds later.

The speed of sound is 340m/s

Homework Equations





The Attempt at a Solution


The distance that covers the speed of sound is the same distance as the one of the fall.

so Ds = Vs(ts) , Df = (1/2)gt^2 also T = tf + ts = 2.50 seconds


I tried to ge rid of ts

ts = (g tf^2) /2Vs then using the ecuation above the quadratic ecuation is

(g tf^2) /2Vs + tf - 2.5

the cuadratic for this is -1 +- (square root of 1) -4(0.01) (-2.5) / 0.02

What have I done wrong here? my solutionary says that tf is 2.42

It initailly divided (g tf^2) /2Vs + tf - 2.5 by Vs I think.
 
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You have a given time - [tex]T[/tex]. The time of the melon falling is [tex]t_1[/tex] and the sound travels [tex]T-t_1[/tex]. Now you have two variables (time and distance) and two equations. Solve these and you get the result.
 
Tom83B said:
You have a given time - [tex]T[/tex]. The time of the melon falling is [tex]t_1[/tex] and the sound travels [tex]T-t_1[/tex]. Now you have two variables (time and distance) and two equations. Solve these and you get the result.

I tried Ds/Vs = T -tf that is (g tf^2) /2Vs + tf - T

I just don't understand why should I divide the quadratic equation by Vs in order to get

(g tf^2) /2 + tf Vs - TVs

Shouldnt (g tf^2) /2Vs + tf - T and (g tf^2) /2 + tf Vs - TVs have the same quadratic result for tf? I have problems only when I try to solve the quadratic equation.
 
Last edited:
Sorry it's to hard to read that. Use TEX and explain what is what...
But as far as I can gather from your posts - you're right that [tex]\frac{Ds}{V_s}=T-t_f[/tex]. For free fall applies [tex]h=\frac{1}{2}gt^2[/tex], in this case [tex]Ds=\frac{1}{2}g{t_f}^2[/tex]. Now you just need to solve these two equations.

Sorry I couldn't answer more precisely to your posts, but I really couldn't understand what you meant. When I look at it again and try to understand it - of course you get two results - one of them should be negative (less than zero), so you just take the result that is positive - you can't have negative time. 2.42 sounds quite correct is the result is supposed to be 2.5