Answer :
The force exerted by the parachutist on the ground is 471.24 N when she comes to rest at a distance of 0.660 m
Let t = Time taken
u = Initial velocity , v = Final velocity
s = Displacement, a = Acceleration
As given in the problem initial velocity u = 3.85 m/s
Final velocity v = 0 (rest)
s = 0.660 m
From the equation of motion v² = u² + 2as
we have a = v²-u² / 2s
⇒a = 0-3.85² / (2* 0.660)
⇒a = -11.22 m/s² (negative sign shows deceleration)
we know force F = mass m * acceleration a
Given mass of parachutist m = 42 kg
∴Force exerted by parachutist on ground
F = 42 * 11.22 = 471.24 N
If the distance is shorter, acceleration increases in magnitude and so does the force exerted by the parachutist.
The question is incomplete and the full question is probably " A 42.0-kg parachutist is moving straight downward with a speed of 3.85 m/s. (a) If the parachutist comes to rest with constant acceleration over a distance of 0.660 m, what force does the ground exert on her? "
Know about when equations of kinematics can be applied
brainly.com/question/14355103
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