You are climbing in the High Sierra when you suddenly find yourself at the edge of a fog-shrouded cliff. To find the height of this cliff, you drop a rock from the top; 7.00 s later you hear the sound of the rock hitting the ground at the foot of the cliff.(a) Ignoring air resistance, how high is the cliff if the speed of sound is 330 m/s? (b) Suppose you had ignored the time it takes the sound to reach you. In that case, would you have overestimated or underestimated the height of the cliff? Explain your reasoning.

When ignoring the time taken by the sound wave to travel up we would have over estimated the height because then we would consider that the whole time was taken by the rock in going down and as we know that distance is calculated as speed time the time taken.

Explanation:

Given:

time delay in the sound being heard after dropping a rock, [tex]t=7\ s[/tex]

speed of sound, [tex]v=330\ m.s^{-1}[/tex]

a)

distance travelled by the rock is equal to the distance travelled by the sound.

Now for rock:

[tex]h=u.t+\frac{1}{2} g.t^2[/tex]

where:

u = initial velocity of the rock = 0

h = height form where the rock is being dropped

[tex]\therefore h=4.9\times t^2[/tex] .....................(1)

when the rock takes time t then the sound will take time (7-t).

∵given time comprises of both the time taken by the rock to reach the ground and the time taken by the sound to travel up.

For sound:

[tex]h=330\times (7-t)[/tex] ......................(2)

Now from (1) & (2)

[tex]4.9\ t^2=2310-330\ t[/tex]

[tex]4.9\ t^2+330\ t-2310=0[/tex]

[tex]t=6.3931\ s[/tex]

b)

When ignoring the time taken by the sound wave to travel up we would have over estimated the height because then we would consider that the whole time was taken by the rock in going down and as we know that distance is calculated as speed time the time taken.