Question

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts help. when she has fallen for 3.0 s , she hears the echo of her shout from the valley floor below. the speed of sound is 340 m/s.

Answer 1

Answer:

$H = 532 m$

Explanation:

When teacher falls into the cliff then she shout for Help at the same time

so here we know that sound will go down and reflect back up

so here in 3 s distance traveled by the sound

$d = vt$

$d = 340 \times 3$

$d = 1020 m$

now in the same time the distance that teacher will fall down is given as

$d_1 = \frac{1}{2}gt^2$

$d_1 = \frac{1}{2}(9.81)(3^2)$

$d_1 = 44.1 m$

now total distance traveled by teacher and sound in 3 s

$d_{total} = d + d_1$

$d_{total} = 1020 + 44.1$

this total distance must be equal to twice the height of the cliff

$2H = 1064.1 m$

$H = 532 m$

Answer 2

Answer:

(a). The height of the cliff from which the teacher falls is $\boxed{532.05\,{\text{m}}}$.

(b). The speed of the teacher just before hitting the ground will be  $\boxed{102.12\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$.

Further Explanation:

Since the teacher listens to the her own sound of shout after $3\,{\text{s}}$, therefore, the sound covers the twice the distance of height of the cliff minus the distance covered by the teacher in  .

Part (a):

The distance covered by the teacher in $3\,\sec$ is:

 $s = ut + \dfrac{1}{2}g{t^2}$    …… (1)                                          

Since the teacher fall from rest, substitute initial velocity to be zero and time to be  .

 $\begin{aligned}s&= \left( {0 \times 3} \right) + \frac{1}{2} \times 9.8 \times {\left( 3 \right)^2} \\&= 44.1\,{\text{m}}\\\end{aligned}$

Write the expression for the height of the cliff.

 $2H - s = v \times t$        …… (2)                                      

Here, $H$ is the height of the cliff, $v$ is the speed of sound and $t$ is the time taken by the sound to travel.

The speed of the sound is $340\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$  and it takes $3\,{\text{s}}$  to reach the teacher.

Substitute the values in the equation (2).

 $\begin{aligned}2H - 44.1 &= 340 \times3\\H&= \frac{{1020 + 44.1}}{2}\,{\text{m}}\\&= {\text{532}}{\text{.05}}\,{\text{m}}\\\end{aligned}$

Thus, the height of the cliff from which the teacher falls is $\boxed{532.05\,{\text{m}}}$.

Part (b):

The teacher falls down the cliff for a height of $532.05\,{\text{m}}$. The speed attained by the teacher as she falls is given by:

$v_f^2 = v_i^2 + 2aH$     …… (3)                                        

The teachers falls with zero velocity under the gravity for a height of $532.05\,{\text{m}}$.

Substitute the values in equation (3).

 $\begin{gathered}v_f^2 &= {\left( 0 \right)^2} + 2\left( {9.8} \right)\left( {532.05} \right)\\{v_f}&= \sqrt {10428.18} \,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right \kern-\nulldelimiterspace} {\text{s}}}\\&=102.12\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{gathered}$

Thus, the speed of the teacher just before hitting the ground will be $\boxed{102.12\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$ .

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Answer Details:

Grade:High School

Chapter:Motion in one dimension

Subject:Physics

Keywords: Physics teacher, outdoor demonstration, high cliff, fallen for 3.0 s, hears echo, valley floor, height of cliff, distance covered, speed of sound, speed of teacher.