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.

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.

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.