Question

A hiker throws a stone directly downward from a cliff with a speed of 10.0 m/sec. The stone travels 35.4 meters downward before hitting a rotten log. What is the speed of the stone just as it reaches the log? Assume there is negligible friction or drag force.
Select one:
a. 28.2 m/sec
b. 29.9 m/sec
c. 24.4 m/sec
d. 26.3 m/sec
e. 33.3 m/sec

Answer 1

Answer on Question #63308, Physics / Mechanics | Relativity

A hiker throws a stone directly downward from a cliff with a speed of $10.0 \, \text{m/sec}$ . The stone travels 35.4 meters downward before hitting a rotten log. What is the speed of the stone just as it reaches the log? Assume there is negligible friction or drag force.

Select one:

a. $28.2\mathrm{m / sec}$

b. $29.9\mathrm{m / sec}$

c. $24.4\mathrm{m / sec}$

d. $26.3\mathrm{m / sec}$

e. $33.3\mathrm{m / sec}$

Solution:

Given:

$y_{0} = h = 35.4\mathrm{m},$

$v_{0x} = 10.0\mathrm{m / s},$

$v_{0y} = 0\mathrm{m / s},$

$v = ?$

The horizontal component of the velocity of the object remains unchanged throughout the motion. The vertical component of the velocity increases linearly, because the acceleration due to gravity is constant $(g = 9.80\mathrm{m / s^2})$

The kinematic equation that describes an object's motion in vertical direction is:

$$v _ {y} ^ {2} = v _ {o y} ^ {2} + 2 g h$$

Thus,

$$v _ {y} = \sqrt {2 g h} = \sqrt {2 \cdot 9 . 8 0 \cdot 3 5 . 4} = 2 6. 3 4 \mathrm {m / s}$$

The horizontal component of velocity is $10.0\mathrm{m / s}$ and the vertical component of velocity is 26.34 m/s.

The final speed is

$$v = \sqrt {v _ {x} ^ {2} + v _ {y} ^ {2}}$$

Thus,

$$v = \sqrt {10 ^ {2} + 26.34 ^ {2}} = 28.2 \mathrm {m / s}$$

Answer: a. 28.2 m/sec

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