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Physics / Physics with MasteringPhysics 4 / Chapter 4 / Problem 17P

Physics with MasteringPhysics | 4th Edition | ISBN: 9780321541635 | Authors: James S. Walker

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A mountain climber jumps a 2.8-m-wide crevasse by leaping

Chapter 4, Problem 17P

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QUESTION:

A mountain climber jumps a 2.8-m-wide crevasse by leaping horizontally with a speed of 7.8 m/s. (a) If the climber's direction of motion on landing is -45°, what is the height difference between the two sides of the crevasse? (b) Where does the climber land?

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QUESTION:

A mountain climber jumps a 2.8-m-wide crevasse by leaping horizontally with a speed of 7.8 m/s. (a) If the climber's direction of motion on landing is -45°, what is the height difference between the two sides of the crevasse? (b) Where does the climber land?

ANSWER:

a.)

Step 1 of 2

We have to find the height difference between the two sides of the crevasse if a mountain climber jumps a 2.8-m-wide crevasse by leaping horizontally with a speed of 7.8 m/s.

The height difference between the two sides of the crevasse is given by the expression

$h-y=$ $\frac{1}{2}$ $g{t}^{2}$

Where,

$t=$ time of leap in s

$g=$ 9.81 m/s

The time of leap can be found using the equation

$t=-$ $\frac{{v}_{x}}{g}$

where,

${v}_{x}=$ horizontal velocity = 7.8 m/s

Thus,

$t=$ $-\frac{7.8}{9.81}$

= 0.8 s

Hence,

The height difference between the two sides of the crevasse is

$h-y=$ $\frac{1}{2}$ $(9.81){(0.8)}^{2}$

= 3.1 m

Therefore, the height difference between the two sides of the crevasse is 3.1 m.

b.)

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