EXAMPLE 4-5 JUMPING A CREVASSE A mountain climber encounters a crevasse in an ice field. The opposite side of the crevasse is 2.74 m lower, and is sepa- rated horizontally by a distance of 4.10 m. To cross the crevasse, the climber gets a running start and jumps in the hori- zontal direction. (a) What is the minimum speed needed by the climber to safely cross the crevasse? (b) If, instead, the climber's speed is 6.00 m/s, where does the climber land? PICTURE THE PROBLEM The mountain climber jumps from x - 0and yo - A - 2.74 m. The land- ing site for part (a) is x - w - 4.10m and y - 0. As for the initial velocity, we are given that co. = to and thy - 0. Finally, with our choice of coordi- nates it follows that a, - 0 and a, - -5. REASONING AND STRATEGY We can model the climber as a projectile, and apply our equations for pro- jectile motion with a horizontal launch. a. From Equations 4-7 we have that x = of and y = 1 = agr. Setting = 0 determines the time of landing. Using this time in the x equa- tion gives the horizontal landing position in terms of the initial speed. b. We can now use the relationship from part (a) to find & in terms of " - 6.00 m/'s. Known Height, A = 2.74 m; width, w = 4.10m. Unknown (a) Minimum initial speed, to - ? (b) Landing distance, x - ?, - W. for vo = 6.00 m/s SOLUTION Part (a) 1. Sety - h - agr equal to zero (landing condition) and solve for the corresponding time : CONTINUED 1157 . Location 3229 of 36046 X W &A Question 25 (1 point) Retake question Please refer to Example 4-5 in the textbook (page 97). What is the speed of the mountain climber at time t = 0.58 s, in meters per second? Keep in mind that the velocity along x does not change. The horizontal velocity is given in the solution to the problem. Calculate the velocity along the vertical direction (Eq. 4-7) and use the Pythagorean Theorem. As always, we use this theorem to find the magnitude of a vector if we know its components