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

[Jimmy84]

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.

 

[Tom83B]

You have a given time - $$T$$. The time of the melon falling is $$t_1$$ and the sound travels $$T-t_1$$. Now you have two variables (time and distance) and two equations. Solve these and you get the result.

 

[Jimmy84]

You have a given time - $$T$$. The time of the melon falling is $$t_1$$ and the sound travels $$T-t_1$$. 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.

 

[Tom83B]

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 $$\frac{Ds}{V_s}=T-t_f$$. For free fall applies $$h=\frac{1}{2}gt^2$$, in this case $$Ds=\frac{1}{2}g{t_f}^2$$. 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

 

[rwisz]

I would like to point out that the solution to this problem involves more than just solving a quadratic equation. There are several factors that need to be considered in order to accurately determine the time it takes for the watermelon to hit the ground.

Firstly, we need to take into account the initial velocity of the watermelon, which is 0. This means that the watermelon is starting from rest and will accelerate due to gravity until it reaches the ground. Secondly, we need to consider the speed of sound, which is the speed at which the sound of the watermelon hitting the ground will travel to the student's ears. This will also affect the time it takes for the student to hear the sound of the impact.

In order to accurately solve this problem, we need to use the equations of motion for constant acceleration, where the initial velocity is 0. These equations are:

1. vf = vi + at
2. d = vit + (1/2)at^2
3. vf^2 = vi^2 + 2ad

where vf is the final velocity, vi is the initial velocity, a is the acceleration (in this case, due to gravity), t is the time, and d is the distance.

Using these equations, we can set up a system of equations to solve for the time it takes for the watermelon to hit the ground. We know that the final velocity of the watermelon is 0 m/s (since it hits the ground and stops moving), the initial velocity is also 0 m/s, and the acceleration due to gravity is 9.8 m/s^2.

Equation 1 becomes 0 = 0 + 9.8t, which simplifies to t = 0 seconds.

Equation 2 becomes d = 0t + (1/2)(9.8)(t^2), which simplifies to d = (4.9)t^2.

We also know that the distance the sound travels is the same as the distance the watermelon falls, so we can set up another equation using the speed of sound (340 m/s).

Equation 3 becomes d = 340t.

Now we have a system of equations:

d = (4.9)t^2
d = 340t

Solving for t, we get t = 0 seconds or t = 69.4 seconds. However, we can disregard the solution