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a student drops a stone from the top of a building and hears the sound of the stone striking the ground after 5.0 s. How high is the building? Assume the speed of sound in air to be 340 m/s.

Sagot :

Glendzmoralesidn

Answer:

The height of the building is 107.5 m

Explanation:

In this problem, we will be using the concept of frree fall and uniform motion to solve for the height of the building. Here, the speed of the air should be considered and not just the pull of gravity.

Let us have the equation.

t(sound) + t(stone) = 5     equation 1

This means that the time for the stone to dropped and the sound it created happened at the same time within 5 seconds.

Solving the problem

a. First, let us apply the free fall formula for the stone using the pull of gravity

y = 1/2gt²

Let us derive this formula to solve the time due to the pull of gravity using cross-multiply

2y = gt²

t(stone) = [tex]\sqrt{2y/g}[/tex]

b. Now, let us use the formula for the uniform motion to create an equation for the stone using the speed of sound

v = d/t

Let us re-write this as

v = y/t     where y is the vertical distance

Again, let us derive this formula to solve for the time due to the speed of sound using cross-multiply

t(sound) = y/v

c. Now, let us substitute the two derived equations to equation 1

t(sound) + t(stone) = 5    

y/v + [tex]\sqrt{2y/g}[/tex] = 5

Simplify the equation

[tex]\sqrt{2y/g}[/tex] = 5 - y/v

2y/g = (5 - y/v)²

Substitute the values: g = 9.8 m/s² (pull of gravity) and v = 340 m/s (speed of sound)

2y/9.8 = (5 - y/340)²

[tex]\frac{2y}{9.8} =25-\frac{10y}{340} +\frac{y^2}{115600}[/tex]

Further, simplify and apply quadratic equation to solve for y

[tex]\frac{y}{4.9} =25-\frac{y}{34} +\frac{y^2}{115600}[/tex]

[tex]\frac{y^2}{115600}-\frac{y}{4.9}-\frac{y}{34} + 25 =0[/tex]

[tex]\frac{y^2}{115600}-\frac{38.9y}{166.6} + 25 =0[/tex]

y = 26884.34 m and y = 107.5 m

Now, since we have two values of y, we need to evaluate this through checking or by substituting each value of y to fit in 5 seconds.

Let us use the formula

t = [tex]\sqrt{2y/g}[/tex]

Then substitute the values of y

t = [tex]\sqrt{2(107.5)/9.8}[/tex] = 4.68 seconds

t = [tex]\sqrt{2(26884.34)/9.8}[/tex] = 74.07 seconds

Therefore, we choose 4.68 seconds < 5 seconds which is more fit for the given time of 5 seconds.

Therefore, the height of the building is 107.5 m

To learn more, just click the following links:

  • Recommendations for free-falling bodies  

       brainly.ph/question/2162209

  • The vertical component of a motion

       brainly.ph/question/2604535

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