Question

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?

Answer 1

Answer:

Vf = 69.61 m/s

Explanation:

We will use the third equation of motion to solve this problem:

$2gh = V_{f}^2 - V_{i}^2$

where,

g = acceleration due to gravity = 9.81 m/s²

h = height of cliff = 247 m

Vf = final velocity = ?

Vi = initial velocity = 0 m/s (boulder breaks loose from rest)

Therefore,

$(2)(9.81\ m/s^2)(247\ m) = V_{f}^2 - (0\ m/s)^2\\V_{f} = \sqrt{4846.14\ m^2/s^2}$

Vf = 69.61 m/s