# Answer to Question #12647 in Mechanics | Relativity for andi

Question #12647

a physics student drops a watermelon from the roof of the building. he hears the watermelon splat 2.50 seconds later. how high is the building? the speed of the sound is 340m/s. ignore air resistance.

Expert's answer

Let's consider the following statement:

T = Tf + Ts,

where T is the total time elapsed from the moment of dropping the watermelon to the moment of hearing the splat (T = 2.5 s), Tf is the flight time of the lemon and Ts is the sound flight time. Let H be the height of a building. Then

H = g·Tf²/2 ==> Tf = √(2H/g)

and

Ts = H / Vs,

where Vs is the speed of the sound (Vs = 340 m/s).

So,

& radic;(2H/g) + H/Vs = T,

or

H/Vs + √H·√(2/g) - T = 0.

We've got a quadratic equation.

& radic;H = ( -√(2/g) ± √(2/g + 4T/Vs) ) · Vs/2 =

= ( -√(2/9.8) ± √(2/9.8 + 4·2.5/340) ) · 340/2 = 5.3478 or -158.9441.

So, the height of a building is H = 5.3478² = 28.5989 m.

T = Tf + Ts,

where T is the total time elapsed from the moment of dropping the watermelon to the moment of hearing the splat (T = 2.5 s), Tf is the flight time of the lemon and Ts is the sound flight time. Let H be the height of a building. Then

H = g·Tf²/2 ==> Tf = √(2H/g)

and

Ts = H / Vs,

where Vs is the speed of the sound (Vs = 340 m/s).

So,

& radic;(2H/g) + H/Vs = T,

or

H/Vs + √H·√(2/g) - T = 0.

We've got a quadratic equation.

& radic;H = ( -√(2/g) ± √(2/g + 4T/Vs) ) · Vs/2 =

= ( -√(2/9.8) ± √(2/9.8 + 4·2.5/340) ) · 340/2 = 5.3478 or -158.9441.

So, the height of a building is H = 5.3478² = 28.5989 m.

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