Suppose the speed of sound is c, and gravity is g. If you throw a rock from an initial height h0 and with an initial velocity v0 the height h(t) of the rock after time t is
h(t)=−g/2 t^2+v0 t+ h0.
(The height is negative when the rock is below ground level. Thus you can think of depth as negative height.)
Suppose you throw the rock down a well and hear the impact after t seconds. Then the depth of this well is d= ?.
(Your answer will be a mathematical expression involving t, g, and c.)
Set up and solve a quadratic equation.
i still don't get it :(
To find the depth of the well, we need to equate the time it takes for the sound to reach us with the time it takes for the rock to hit the bottom.
Let's consider the time it takes for the sound to reach us first. The distance traveled by sound is equal to the speed of sound multiplied by the time taken. Since the sound is traveling upwards, the distance covered by sound would be the negative of the depth of the well, denoted as "d". Hence, the equation for the distance covered by sound is:
-distance covered by sound = -c * t
Now, let's consider the time it takes for the rock to hit the bottom of the well. The height of the rock at time t is given by h(t) = -g/2 * t^2 + v0 * t + h0. When the rock hits the bottom, its height is 0. We can set h(t) = 0 and solve for the time taken:
0 = -g/2 * t^2 + v0 * t + h0
This equation is quadratic in nature. We can use the quadratic formula to solve for t. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
In our case, the coefficients are: a = -g/2, b = v0, and c = h0.
Applying the quadratic formula to our equation, we have:
t = (-v0 ± sqrt(v0^2 - 4*(-g/2)*h0)) / (2*(-g/2))
Simplifying further:
t = (-v0 ± sqrt(v0^2 + g * h0)) / (-g)
Now, we equate the time taken for the sound and the time taken for the rock to hit the bottom:
-c * t = -d
Substituting the expression for t, we have:
-c * ((-v0 ± sqrt(v0^2 + g * h0)) / (-g)) = -d
Canceling out -g and -c on both sides:
(d/c) = ((-v0 ± sqrt(v0^2 + g * h0)) / g)
The depth of the well, d, is given by the expression ((-v0 ± sqrt(v0^2 + g * h0)) / g), where v0 is the initial velocity of the rock, g is the acceleration due to gravity, and h0 is the initial height of the rock.
The time it takes for the sound to come back is d/c
The time it takes to fall a distance of d is found using
h0 + v0 t - g/2 t^2 = d
Solve that for t, and then you have
t = 1/g (v0 + √(v0^2 - 2g(d-h0)))
Now you know that
1/g (v0 + √(v0^2 - 2g(d-h0))) + d/c = t
Now just solve that for d.