A box slides across a frictionless floor with an initial speed of 1.8m/s. It encounters a rough region where the coefficient of frictionless is 0.7. What is the shortest length of rough floor will stop the box? V=1.8 uk=0.7

answer: the correct answer is   air pressure at the top of mount everest is higher than at the bottom of the grand canyon.

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explanation:

answer; how far out of place which the object finished relative to the starting location;

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initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft