|A 80.0 kg runner is looking to jump off an icy cliff (us = 0.20, µg = 0.10). The runner startsaccelerating from rest 2.50 m from the edge before jumping off horizontally.8.What is the maximum acceleration the runnerа.can have before his feet slip on the ground?2.50 mb.|Given that acceleration, does the runner landon the ledge or in the water below? Show using yourcalculations.6.00 m1.75 mLedge

a. The maximum acceleration will be such a value so that the force acting on the runner equal to kinetic frictional force. Therefore if we assume acceleration is "a" mass is m, then ma= μkmg Here acceleration due to gravity g=9.8m/s2 Or a=μkg Putting values we get a= 0.10x9.8=0.98m/s2 This is the maximum acceleration.b) first we need to determine the runner's speed at the edge after traveling 2.5m From third equation of motion v2 = u2 + 2as v2 = 0 + 2x0.98x2.5=4.9 Or Now when the runner jump off at the edge his the acceleration in vertical direction will be equal to g. And his initial vertical velocity will be zero. Therefore from second equation of motion time taken to travel a vertical distance s=6.00m is In this time the runner can move distance in horizontal direction will be again from second equation of motion ( remember here initial speed u=) Therefore he will fall in the water, since (2.45>1.75).

Answered: A 80.0 kg runner is looking to jump off…