Q78P - University Physics with Modern Physics

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. (a) how tall is the cliff? (b) If we ignore air resistance, how fast will she be moving just before she hits the ground?

Step-by-Step Solution

Answer

a)  Height of the cliff is 532 m.

b)  Speed is 14.7 m/s .

Step 1: Identification of given data

Fallen time (t) = 3.0 s

Speed (v) = 340 m/s

Step 2: Calculation for the height

As we know that

$$\begin{gathered} d=vt \\ d=340\times 3 \\ d=1020m \end{gathered}$$

Now in the same time teacher will fall down that distance,

 $$\begin{gathered} d_1=\frac{1}{2}g{t}^2 \\ {d}_1=\frac{1}{2}\times 9.81\times 9 \\ {d}_1=44.1 m \end{gathered}$$

Now total distance travelled by teacher and sound

 $$\begin{gathered} d_t=d_1+d \\ {d}_t=1020+44.1 \\ {d}_t=1061.1m \end{gathered}$$

This distance will be equal to twice of the height of cliff

 $$\begin{gathered} 2H=1064.1 \\ h=532m \end{gathered}$$

Step 3: Calculation for the speed

$$\begin{gathered} d_1=vt \\ 44.1=v\times 3 \\ v=14.7m/s \end{gathered}$$