The total time it takes the sound to reach our on top of the cliff (10.0 s) is actually sum of two times:
- the time it takes the rock to hit the ground starting from the top of the cliff, t1
- the time it takes the sound to reach the top of the cliff starting from the ground, t2
1) The rock moves by uniformly accelerated motion, with constant acceleration
. Its law of motion is given by 
, where 
is the initial velocity of the rock, and h is the height of the cliff. The time t1 is the time at which the rock reaches the ground, so that y(t1)=0, and the equation becomes
2) The sound moves from the ground to the top of the cliff by uniform motion, with constant speed v=343 m/s. Therefore, the sound covers the distance h (the height of the cliff) in a time t2 given by
3) If we rewrite h in both equations, we can write:
(1)
4) We also know that the sum of t1 and t2 is equal to 10 seconds:
from which we find
if we substitute this into eq.(1), we get
Numerically:
Solving the equation, we find the solution
(the other solution is negative, so it does not have physical meaning). As a consequence,
and the height of the cliff is given by
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- the time it takes the rock to hit the ground starting from the top of the cliff, t1
- the time it takes the sound to reach the top of the cliff starting from the ground, t2
1) The rock moves by uniformly accelerated motion, with constant acceleration
2) The sound moves from the ground to the top of the cliff by uniform motion, with constant speed v=343 m/s. Therefore, the sound covers the distance h (the height of the cliff) in a time t2 given by
3) If we rewrite h in both equations, we can write:
4) We also know that the sum of t1 and t2 is equal to 10 seconds:
from which we find
if we substitute this into eq.(1), we get
Numerically:
Solving the equation, we find the solution
and the height of the cliff is given by
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