Respuesta :
Answer:
[tex]H = 532 m[/tex]
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
[tex]d = vt[/tex]
[tex]d = 340 \times 3[/tex]
[tex]d = 1020 m[/tex]
now in the same time the distance that teacher will fall down is given as
[tex]d_1 = \frac{1}{2}gt^2[/tex]
[tex]d_1 = \frac{1}{2}(9.81)(3^2)[/tex]
[tex]d_1 = 44.1 m[/tex]
now total distance traveled by teacher and sound in 3 s
[tex]d_{total} = d + d_1[/tex]
[tex]d_{total} = 1020 + 44.1[/tex]
this total distance must be equal to twice the height of the cliff
[tex]2H = 1064.1 m[/tex]
[tex]H = 532 m[/tex]
Answer:
(a). The height of the cliff from which the teacher falls is [tex]\boxed{532.05\,{\text{m}}}[/tex].
(b). The speed of the teacher just before hitting the ground will be [tex]\boxed{102.12\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}[/tex].
Further Explanation:
Since the teacher listens to the her own sound of shout after [tex]3\,{\text{s}}[/tex], 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 [tex]3\,\sec[/tex] is:
[tex]s = ut + \dfrac{1}{2}g{t^2}[/tex] …… (1)
Since the teacher fall from rest, substitute initial velocity to be zero and time to be .
[tex]\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}[/tex]
Write the expression for the height of the cliff.
[tex]2H - s = v \times t[/tex] …… (2)
Here, [tex]H[/tex] is the height of the cliff, [tex]v[/tex] is the speed of sound and [tex]t[/tex] is the time taken by the sound to travel.
The speed of the sound is [tex]340\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}[/tex] and it takes [tex]3\,{\text{s}}[/tex] to reach the teacher.
Substitute the values in the equation (2).
[tex]\begin{aligned}2H - 44.1 &= 340 \times3\\H&= \frac{{1020 + 44.1}}{2}\,{\text{m}}\\&= {\text{532}}{\text{.05}}\,{\text{m}}\\\end{aligned}[/tex]
Thus, the height of the cliff from which the teacher falls is [tex]\boxed{532.05\,{\text{m}}}[/tex].
Part (b):
The teacher falls down the cliff for a height of [tex]532.05\,{\text{m}}[/tex]. The speed attained by the teacher as she falls is given by:
[tex]v_f^2 = v_i^2 + 2aH[/tex] …… (3)
The teachers falls with zero velocity under the gravity for a height of [tex]532.05\,{\text{m}}[/tex].
Substitute the values in equation (3).
[tex]\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}[/tex]
Thus, the speed of the teacher just before hitting the ground will be [tex]\boxed{102.12\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}[/tex] .
Learn More:
- The amount of kinetic energy an object has depends on its https://brainly.com/question/947434
- The cheetah can maintain its maximum speed for only 7.5 s. What is the minimum distance the gazelle must be ahead https://brainly.com/question/11270989
- A car's position in relation to time is plotted on the graph. What can be said about the car https://brainly.com/question/11314764
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.
