Physics, 05.02.2021 21:30 bionicboy03120440

A building inspector standing on the top floor of a building wishes to determine the depth of the elevator shaft. They drop a rock from rest and hear it hit bottom after 2.56 as. (a) How far (in m) is it from where they drop the rock to the bottom of the shaft

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Answer from: jrassicworld4ever

d = 29.89 m

Explanation:

To solve this, we need to separate this problem in two parts.

One part would be the the time taken by the rock to actually hit the bottom, and the other part would be the time taken by the sound to reach the inspector.

Joining these two times we have:

t = t₁ + t₂ (1)

This time is 2.56 s.

Now, as we are asked to determine the distance from the top floor to the bottom, and we have two times taken in different ways, one by sound and the other the actual, we can say the same thing on distance, we need a distance relationed to the time taken by rock to hit the bottom, and the other distance relationet to the time taken by sound to reach the inspector.

Doing this we have that the distance traveled by the rock is:

y₁ = gt²/2

y₁ = 9.8t²/2 = 4.9t₁² (2)

Now, the distance traveled by sound would be:

y₂ = v * t₂ = 336t₂ (3)

Remember that the speed of the sound is 336 m/s

From this last expression (3), we can actually write t₂ in function of t₁, using (1):

2.56 = t₁ + t₂

t₂ = 2.56 - t₁ (4)

Replacing (4) in (3):

y₂ = 336(2.56 - t₁) (5)

Now that we have y₁ and y₂, we can equal (2) and (5), both expressions to get the value of t₁, and then, calculate the distance:

4.9t₁² = 336(2.56 - t₁)

4.9t₁² = 860.16 - 336t₁

4.9t₁² + 336t₁ - 860.16 = 0

Using the quadractic formula, we can calculate t₁:

t₁ = -336 ±√(336)² + 4*4.9*860.16 / (2*4.9)

t₁ = -336 ±√129,7555.136 / 9.8

t₁ = -336 ± 360.21 / 9.8 Using only the positive value we have:

t₁ = 2.47 s

This means that the rock hits the bottom in 2.47 s, and the remaining 0.09 s belongs to the time taken by sound. (2.47 + 0.09 = 2.56 s)

With this, we can calculate the distance of the rock using expression (2):

y₁ = 4.9 * (2.47)²

y₁ = 29.89 m

Hope this helps

Answer from: Quest

Is a vector quantity that refers to "the rate at which an object changes its position." imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. while this might result in a frenzy of activity, it would result in a zero velocity. because the person always returns to the original position, the motion would never result in a change in position. since velocity is defined as the rate at which the position changes, this motion results in zero velocity. if a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position. every step must go into moving that person further from where he or she started. for certain, the person should never change directions and begin to return to the starting position.

Answer from: Quest

prndulum needs thread, weight clock to time swings

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