As Wile E. Coyote (mass = 21 kg) ages, he finds it harder to chase the roadrunne

Question

As Wile E. Coyote (mass = 21 kg) ages, he finds it harder to chase the roadrunner. He goes on eBay and finds a pair of frictionless, rocket-powered roller skates from the Acme Company. When they arrive, he puts on the skates with great excitement and fires them up He finds that they provide a constant thrust of 315 N in a horizontal direction Off he goes chasing the roadrunner The roadrunner leads the coyote to the edge of a cliff, then escapes by making a sudden turn. The unfortunate coyote finds himself 10 m from the edge of the cliff, and because of his large inertia, is unable to turn To his horror, the controls of the skates jam, and they continue operating with no reduction in acceleration. At that instant he is traveling at 30 m/s, and continues forward off the edge of the cliff, which is a distance of 100 m above the desert floor. If the coyote's skates remain operating and horizontal, so that they still provide the horizontal acceleration, how far from the base of the cliff does the coyote land? What are the horizontal and vertical components of his velocity the instant before he crashes into the desert floor? If the coyote's impact with the desert floor brings him to rest in a time of 3.0 ms, what force does the desert floor exert on the coyote?

Explanation / Answer

acceleration provided by the thrust=force/mass=315/21=15 m/s^2

let horizontal speed just before dropping from the edge=v

then using the formula:

final velocity^2-initial velocity^2=2*acceleration*distance

v^2-30^2=2*15*10

==>v=sqrt(30^2+2*15*10)=34.641 m/s

his initial vertical speed=0 m/s

vertical acceleration=9.8 m/s^2

then time taken to reach the ground=sqrt(2*height/acceleration)=sqrt(2*100/9.8)=4.5175 seconds

then horizontal distance from the edge of the cliff=initial horizontal speed*time+0.5*horizontal acceleration*time^2

=34.641*4.5175+0.5*15*4.5175^2=309.55 m

horizontal component of veloicty before hitting floor=initial veloicty+horizontal acceleration*time

=34.641+15*4.5175=102.4 m/s

vertical component=initial vertical speed+vertical acceleration*time

=0+9.8*4.5175=44.272 m/s

total speed=sqrt(102.4^2+44.272^2)=111.56 m/s

let force applied is F.
then impulse applied=F*time =F*0.003 N.sec

as we know, impulse applied=change in momentum=mass*(final speed-initial speed)=21*(0-111.56)

magnitude of change in momentum=21*111.56=2342.8 kg.m/s

hence 0.003*F=2342.8

==>F=7.8092*10^5 N

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