2-D physics question?

Question

If a rock takes 7.5 seconds for it to fall, hit the ground, the the sound to bounce back up, what is the height it fell from? If someone could explain to me how to solve this (and dumb it down, because I don't understand physics at all...) THANKS.

acceleration due to gravity is -9.8 meters per second squared, and the speed of sound is 340.29 meters per second.

Answer 1

first identify the motion of the rock and sound.

the motion of the rock is an acceleration
the motion of the sound is constant

Xf = 1/2at^2 + Vt + Xi (for rock)
Xf = Vt + Xi (for sound)

Xf = final position
Xi = initial position
a = acceleration
t = time
V = velocity

The rock is released from an initial height (Xi), it falls to its final height (Xf), which is 0m. At the final height of the rock, the sound starts to travel, and it travels from the final height of the rock to the inital height of the rock.

Notice that the intial height of the rock is the final postion of sound. Thus, Xi (rock) = Xf (sound)

from these two equations:
Xf = 1/2at^2 + Vt + Xi (for rock)
Xf = Vt + Xi (for sound)

Now list everything we know, and get rid of unnessessary information.

Since the rock falls from rest, we don't need Vt part of the equation. So we have:

Xf = 1/2at^2 + Xi
Xf = Vt + Xi

We know the final height of the rock, initial position of sound, speed of sound and acceleration.

Final height of the rock: 0m
Initial position of sound: 0m

Plug the numbers in:
0 = 1/2(-9.8)t^2 + Xi
Xf = 340.29t + 0

Now let t be the time it takes the rock to hit the groud, then 7.5 - t is the time it takes sound to reach back to the intial height of the rock

Again, final postion of sound is the initial height of the rock. From this information, plug Xf of sound for Xi of rock

0 = 1/2(-9.8)t^2 + 340.29(7.5 - t) + 0

0 = -4.9t^2 + 340.29(7.5 - t)

0 = -4.9t^2 + 2552.175 - 340.29t

use quadratic formula and you'll get t = 6.8285627s

use the time to find the height
0 = 1/2(-9.8)(6.8285627)^2 + Xi
Xi = 228.48m

The rock is released from 228.48m from the ground

hope this helps

Answer 2

If you drop the rock from rest, i.e. don't throw it, the time it takes to reach the bottom is sqrt(2d/g). The time it takes for the sound to come up to you is d/c, where c = the speed of sound. The total time is the sum of these two times. Adding them together and solving for d gives the distance . . .

Answer 3

sqrt(2d/a) + (d/c) = t

(2d/a) + (d/c)^2 = t^2

(2dc^2) + (ad^2) =a(tc)^2

(a)d^2 + (2c^2)d = a(tc)^2

you know the numbers for acceleration a(or g) which is 9.8, and c which is the speed of sound, they give you the time so just solve the algebra for d and that is your distance. you may have to use the quadratic equation if you dont have a fancy calculator like a TI 89.