Respuesta :
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 [tex]g=9.81 m/s^2[/tex]. Its law of motion is given by [tex]y(t)=h-v_0t-\frac{1}{2}gt^2[/tex], where [tex]v_0=0[/tex] 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
[tex]0=h-\frac{1}{2}gt_1^2[/tex]
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
[tex]h=vt_2[/tex]
3) If we rewrite h in both equations, we can write:
[tex]=\frac{1}{2}gt_1^2=vt_2[/tex] (1)
4) We also know that the sum of t1 and t2 is equal to 10 seconds:
[tex]t_1+t_2=10[/tex]
from which we find
[tex]t_2=10-t_1[/tex]
if we substitute this into eq.(1), we get
[tex]\frac{1}{2}gt_1^2+vt_1-10v=0[/tex]
Numerically:
[tex]4.9t_1^2 + 343 t_1 - 3430=0[/tex]
Solving the equation, we find the solution [tex]t_1=8.87 s[/tex] (the other solution is negative, so it does not have physical meaning). As a consequence,
[tex]t_2 = 10-t_1 = 1.13 s[/tex]
and the height of the cliff is given by
[tex]h=vt_2=(343 m/s)(1.13 s)=388 m[/tex]
- 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 [tex]g=9.81 m/s^2[/tex]. Its law of motion is given by [tex]y(t)=h-v_0t-\frac{1}{2}gt^2[/tex], where [tex]v_0=0[/tex] 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
[tex]0=h-\frac{1}{2}gt_1^2[/tex]
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
[tex]h=vt_2[/tex]
3) If we rewrite h in both equations, we can write:
[tex]=\frac{1}{2}gt_1^2=vt_2[/tex] (1)
4) We also know that the sum of t1 and t2 is equal to 10 seconds:
[tex]t_1+t_2=10[/tex]
from which we find
[tex]t_2=10-t_1[/tex]
if we substitute this into eq.(1), we get
[tex]\frac{1}{2}gt_1^2+vt_1-10v=0[/tex]
Numerically:
[tex]4.9t_1^2 + 343 t_1 - 3430=0[/tex]
Solving the equation, we find the solution [tex]t_1=8.87 s[/tex] (the other solution is negative, so it does not have physical meaning). As a consequence,
[tex]t_2 = 10-t_1 = 1.13 s[/tex]
and the height of the cliff is given by
[tex]h=vt_2=(343 m/s)(1.13 s)=388 m[/tex]
Answer:
height of the cliff would be
H = 385.25 m
Explanation:
here when we drop the stone from the cliff then first stone will hit the rock and then the sound of hitting the rock will come to us.
So here this total time is given as t = 10 s
now we know here that time to fall the stone is given by kinematics
[tex]y = \frac{1}{2}gt^2[/tex]
[tex]t = \sqrt{\frac{2H}{g}}[/tex]
now the time to travel the sound from bottom to us is given as
[tex]t' = \frac{H}{340}[/tex]
here we assume that speed of sound is 340 m/s
now we have
[tex]t + t' = 10 s[/tex]
[tex]\sqrt{\frac{2H}{g}} + \frac{H}{340} = 10[/tex]
by solving above equation we have
H = 385.25 m
