M1 newtons laws question? cant solve it....:S help please?

(a)

Falling from rest, the distance d fallen after t seconds is given by

d = ½gt²

Solving for t,

t = √(2d/g)

If d = 10 m, then t = √(20/g) = 2 √(5/g) sec. Put in whatever value for g (in m/sec²) you're supposed to use (10 m/s², 9.8 m/s², etc.)

At this time, the person will reach a speed v given by

v = gt = √(20g) = 2 √(5g)

(b)

Assuming that the rope doesn't stretch, the instant the rope becomes taut, the 200kg mass is lifted infinitesimally off the floor. This means that the normal force no longer acts on it. At this moment, the tension in the rope replaces the normal force, so the tension in the rope is 200g newtons.

Now consider the falling person. Until this moment he has been acted on only by gravity, but now he is acted on by the tension in the rope as well, so at this moment the force on him is(200 - 70)g = 130g N upward. The acceleration at that moment (and thereafter) is 130g/70 = 13g/7 newtons, upward.

(c)

Here I will use the equation

(v_f)² = (v_i)² + 2ad

where v_f is the final velocity, v_i is the initial velocity, a is the constant force, and d is the distance traveled during the time that the velocity changed from v_i to v_f.

Here we take v_f = 0, v_i = 2 √(5g), a = -13g/7

You can solve for d.

This was a challenging problem!