(II) Extreme-sports enthusiasts have been known to jump off the top of El Capita
ID: 1865079 • Letter: #
Question
(II) Extreme-sports enthusiasts have been known to jump off the top of El Capitan, a sheer granite cliff of height 910 m in Yosemite National Park. Assume a jumper runs horizontally off the top of El Capitan with speed 4.0 m/s and enjoys a free fall until she is 150 m above the valley . (II) Three vectors are shown in Fig. 3-35. Their magnitudes floor, at which time she opens her parachute (Fig. 3-37). are given in arbitrary units. Determine the sum of the three (a) How long is the jumper in free fall? Ignore air resis- vectors. Give the resultant in terms of (a) components, tance. (b) It is important to be as far away from the clif (b) magnitude and angle with the +x axis. as possible before opening the parachute. How far from the cliff is this jumper when she opens her chute? 4.0 m/s 44.0) 56. A (A 28.0° FIGURE 3-35 Problems 9, 10, 11, 12, and 13 Vector magnitudes are given in arbitrary units ? (C-31.0) 10 m 150 m FIGURE 3-37 blem 26.Explanation / Answer
1)
Given vectors are
A = 44 cos28 i + 44 sin 28 j
B = 26.5 cos(180-56) i+ 26.5 sin(180-56) j
C = 31 cos 270 i + 31 sin 270 j
resultant vector is
R = A+B+C
R = (Rx) i + Ry j
R = (Ax+Bx+Cx)i + (Ay+By +Cy) j
R = (44 cos28 + 26.5 cos(180-56)+31 cos 270)i + ((44 sin28 + 26.5 sin(180-56)+31 sin 270)) j
R = (38.85 -14.82 +0) i+(20.66+21.97-31)
R = (24.03) i + (11.63) j
the magnitude is
R = sqrt(24.03^2+11.63^2) = 26.7
the direction is
theta = arc tan (Ry/Rx)
= arc tan (11.63/24.03)
= 25.82 degrees
from +x axis
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2)
Given jumper initially at a height of h = 910 m
jumps horizontally with initial velocity ux = 4 m/s
opens paachute at 150 m from the floor
a) from equations of motions
s = ut +0.5*a*t^2
y-y0 = v0y*t -0.5*g*t^2
y= y0 - v0y*t -0.5*g*t^2
here voy=0 m/s , y=h
h = y0-0.5*g*t^2
solving for t
t = sqrt(2(y0-h)/g)
substituting the values
t = sqrt(2(910-150)/9.8)s
t = 12.454 s
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b) how far the jumper from the cliff before opening the parachute is
V = x/t ==> x = v*t
here v is ux and t is time at which jumper opens the parachute
x = 4*12.454 m = 49.816 m
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