There is a 247–m–high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff. What is the velocity of the boulder just before it strikes the ground?

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.Explanation:

dang that balls went far