What is the force required to stop a 20 kg man traveling at 12.3 m/s east in 0.9 s? *round to two decimal points and include your direction! *

F = ma

We have mass = 20kg

And we need to solve for acceleration

So acceleration is change in velocity over time, in this case we have one velocity and we can assume the man started from rest so

12.3 / 0.9 = a

a = 13.6667

Now we can plug that into F = ma

F = (20)(13.6667)

F = 273.334

Rounding

F = 273.33

Now he is traveling east so we need a force towards the rest, or in the opposite direction to stop his motion.

If we assume east is the positive direction then we need a force of

-273.33 N to stop the man or 273.33 towards the west.

where is the photo and the answer choices?

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