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.

How fast will the boulder be going when it strikes the ground?
1. v_f = sqrt{2gh}
2. v_f = frac{2gh}{t}
3. v_f = frac{gt}{2h}
4. v_f = frac{2h}{gt}

Answer 1

The boulder will be going approximately 70.0 m/s when it strikes the ground. The tourist will have approximately 0.7463 seconds to get out of the way after hearing the sound of the rock breaking loose. Therefore, the correct answer is 1. v_f = sqrt{2gh}.

Step-by-step explanation:

To find the speed at which the boulder will be going when it strikes the ground, we can use the equation:

vf = sqrt(2gh)

where vf is the final velocity, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the cliff (250 m).

Plugging in the values, we get:

vf = sqrt(2 * 9.8 * 250) = sqrt(4900) = 70.0 m/s

Therefore, the boulder will be going approximately 70.0 m/s when it strikes the ground.

To determine how long the tourist at the bottom will have to get out of the way after hearing the sound of the rock, we can calculate the time it takes for the sound to reach the tourist:

time = distance / speed

The distance is the height of the cliff (250 m) and the speed is the speed of sound (335 m/s). Plugging in the values, we get:

time = 250 / 335 = 0.7463 s

Therefore, the tourist will have approximately 0.7463 seconds to get out of the way after hearing the sound of the rock breaking loose.