<span>I believe this question has additional detail which stated
that during the 1st half, his speed was 2.01 m/s. From this we can
calculate his speed during the second half, v2, using the formula:</span>
v_ave = (v1 + v2) / 2
2.05 m/s = (2.01 m/s + v2) / 2
<span>v2 = 2.09 m/s</span>
Looking at the force-time graph, wouldn't the force be integral fdt between 0 and 10s, a sort of "smoothed out pulse" ? And it looks like a familiar bell shaped curve. doesn't that produce a turning effect/torque and isn't there something about a circular analogue to F=ma in newtonian linear mechs ???
Looks like a difficult question for 5 points
Answer:

Explanation:
Impulse-momentum theorem states that impulse is equal to the change of momentum:
(1)
with pf the final momentum and pi the initial momentum. Knowing that momentum is mass (m) times velocity (1) is:

It's important to note that we're dealing with vector quantities so direction matters. If we choose towards the floor positive direction then the initial velocity is positive and the final velocity is negative, so:


So, the impulse delivered to the floor is 
Given Information:
Acceleration = a = 3.5×10⁵g
Gravitational constant = g = 9.8 m/s²
Distance from axis of rotation = r = 2.0 cm = 0.02 m
Required Information:
Angular speed = ω = ?
Answer:
Angular speed = ω = 1.31×10⁴ rad/s
Explanation:
We know that the centripetal acceleration is given by
a = v²/r
The relationship between linear and angular speed is given by
v = ωr
Where ω is the angular speed
So the acceleration becomes
a = (ωr)²/r
a = ω²r²/r
a = ω²r
Re-arrange for ω
ω² = a/r
ω = √(a/r)
The acceleration is
a = 3.5×10⁵g
where g is the gravitational constant
a = 3.5×10⁵*9.81
a = 3.434×10⁶ m/s²
Finally, the required angular speed is
ω = √(3.434×10⁶/0.02)
ω = 1.31×10⁴ rad/s