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 a time 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 335.0 m/s on this day.

a) = 70 m/sec   b= 6.1 sec.

Explanation:

If the boulder "breaks loose" this means that it starts from rest, so vo=0.

So, as we know the height from which the boulder is falling (250 m), and assuming that the acceleration of the object is due to gravity, we can write the following:

h= 1/2 g t²= 250 m ⇒t= 7.14 sec.

By definition, we know that as a=g, and a= vf-vo/t, we can obtain vf (the speed with which the object strikes ground) as follows:

vf= gt ⇒ 9.8 m/s. 7.14 sec = 70 m/sec

a) = 70 m/sec

Now, in order to know the time needed for the tourist to keep safe, we need to know first, how long after the rock broke, he listened the noise.

Assuming that the sound speed is constant and equal to 335.0 m/s, we can apply the definition of velocity, and solve for time as follows:

ts = h / vs = 250 m / 335.0 = 0.74 sec.

If we add the reaction time of 0.3 sec, we get t= 1.04 sec.

As we have already know that the time at which the rock will strike ground, will be 7.14 sec after breaking, we can conclude that the tourist must get out of the way before 6.1 sec. after reacting to the noise of the rock falling in order to avoid being hit by the object.