Question

10:59 PM otion A mountain climber encounters a crevasse in an ice field. The opposite side of the crevasse is 2.75 m lower, and is separated horizontally by a distance of 4.10 m. To cross the crevasse, the climber gets a running start and jumps in the horizontal direction. a) If the climber's initial speed is 6.00 m/s, where does the climber land and what is the speed on landing? What is the minimum speed needed by the climber to safely cross the crevasse? b) TARI FTON
Doing test review. Checking to see if my answers correct.

Answer 1

Given,

h = 2.75 m ; D = 4.1 m ;

a)vi = 6 m/s

the time taken will be:

t = sqrt (2h/g) = sqrt (2 x 2.75/9.8) = 0.75 s

D = vi T = 6 x 0.75 = 4.5 m

the horizontal velocity remains same but the vertical velocity at landing is:

vy = sqrt (2 g h) = sqrt (2 x 9.8 x 2.75) = 7.34 m/s

v = sqrt (6^2 + 7.34^2) = 9.48 m/s

Hence, D = 4.5 m ; v = 9.48 m/s

b)let min speed be v,

we have, t = 0.75 s (calculated i (a))

D = v t => v = dx/t = 4.1/0.75 = 5.47 m/s

Hence, v = 5.47 m/s