Q51PE - College Physics (Urone)

Question

Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can’t see the rock right away but then does,  1.50 s later. 

(a) How far above the hiker is the rock when he can see it? 

(b) How much time does he have to move before the rock hits his head?

Step-by-Step Solution

Answer

a) 93.97 m

b) 3.127 s

Step 1: Given data for a free-falling object

• The stone is moving downwards; hence the above is the case of a free-falling body.
• The height from where the rock breaks loose is 105 m  .
• The person can see it after 1.50 s  .

Step 2: Velocity and distance

a)

Data available :

• Initial velocity: U=0.
• Time: T= 1.50 s.
• Acceleration: g=9.81 m/s².

The distance traveled by the rock can be calculated from the equation of motion as :

$\mathbf{d}\mathbf{=}\mathbf{Ut}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{gt}}^{\mathbf{2}}$ ………… (1)

Here d is the distance, U is the initial velocity, g is the constant, and t is the time taken by the object to travel the distance.

Substituting values in the above equation, we get:

$\begin{array}{rcl}\mathrm{d}& =& \left(0\right)\times \left(\mathrm{t}\right)+\frac{1}{2}\times \left(9.81\right)\times {\left(1.5\right)}^2\\ & =& 11.03 \mathrm{m}\end{array}$

The distance traveled by the rock is part a 11.03 m  .

We also need to find out the velocity at the interface point, and that can be calculated as:

$\mathbf{V}\mathbf{=}\mathbf{U}\mathbf{+}\mathbf{gt}$

 Substituting values in the above equation, we get:

$\begin{array}{rcl}\mathrm{V}& =& 0+\left(9.81\right)\times \left(1.5\right)\\ & =& 14.71 \raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.\end{array}$

 

Hence the velocity at the point of the interface from where a person starts seeing rocks is 14.71 m/s.

Step 3: Distance above the hiker

The distance from where the person can see the rock is very easily found

$$\begin{gathered} d=105 m-11.03 m \\ d=93.97 m \end{gathered}$$

 

Hence the distance from where he can see the rock is 93.97 m.

Step 4: Time he has to move before the rock hits his head

b)

Data available for part b:

• U= 14.71 m/s
• g=9.81 m/s
• d= 93.97 m

From equation (1), we can calculate the required time as:

$$\begin{gathered} 93.97=\left(14.71\right)\times \left(t\right)+\frac{1}{2}\times \left(9.81\right)\times {\left(t\right)}^2 \\ 4.905t^2+14.71t-93.97=0 \end{gathered}$$

 

The root of the above calculation that makes sense is 3.127 s  .

The time he has to move before the rock hits his head is 3.127 s  .