Question

There is a 250-m-high cliff at half dome in yosemite national park in california. suppose a boulder breaks loose from the top of this cliff. (a) how fast will it be going when it strikes the ground? (b) assuming a reaction time of 0.300 s, how long will a tourist at the bottom have to get out of the way after hearing the sound of the rock breaking loose (neglecting the height of the tourist, which would become negligible anyway if hit)? the speed of sound is 335 m/s on this day.

Answer 1

Part A. For this part, we use two equations for linear motion:

y = y0 + v0 t + 0.5 g t^2                   ---> 1

vf = v0 + g t                                         ---> 2

First we solve for t using equation 1: y0 = 0 (initial point at top), y = 250 m, v0 = 0 (at rest)

250 = 0.5 (9.8) t^2

t = 7.143 s

Now we solve for final velocity vf using equation 2:

vf = g t

vf = 9.8 (7.143)

vf = 70 m/s

 

Part B. First we solve for the time it takes for the sound to reach the tourist.

t(sound) = 250 / 335 = 0.746 s

Therefore the total time would be:

t = 0.746 s + 0.300 s

t = 1.05 s

 

Hence there is enough time for the tourist to get out before the boulder hits him.

Answer 2

(a) It will strike the ground in 7.14 seconds

(b) The tourist at the bottom must get out in 6.10 seconds

Further explanation

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem !

Given:

h = 250 m

Unknown:

t₁ = ?

t₂ = ?

Solution:

Question (a) :

We can use the following formula to calculate time taken for the rock to reach the ground,

$h = v ~ t + \frac{1}{2} ~ g ~ t^2$

$250 = 0 \times t + \frac{1}{2} \times 9.8 ~ t^2$

$250 = \frac{1}{2} \times 9.8 ~ t^2$

$250 = 4.9 ~ t^2$

$t^2 = 250 \div 4.9$

$t = \sqrt{250 \div 4.9}$

$t = \frac{50}{7} ~ seconds$

$\large {\boxed {t \approx 7.14 ~ seconds} }$

Question (b) :

Firstly, we will find the time taken by sound to reach the tourist.

$t_{sound} = \frac{250 ~ m}{335 ~ m/s}$

$\boxed{ t_{sound} = \frac{50}{67} ~ s }$

Next, we will find how long will a tourist have to get out of the way.

$t_2 = t_1 - ( t_{sound} + t_{reaction})$

$t_2 = \frac{50}{7} - ( \frac{50}{67} + 0.300})$

$\large {\boxed {t_2 \approx 6.10 ~ seconds} }$

The tourist only has about 6 seconds before getting hit by the stone.

Learn more

• Velocity of Runner : brainly.com/question/3813437
• Kinetic Energy : brainly.com/question/692781
• Acceleration : brainly.com/question/2283922
• The Speed of Car : brainly.com/question/568302

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate