Answer:
What is the average acceleration of the man from the time his feet first touch the patio to the time he is brought fully to rest?
168.97m/s/s
With what force does this jump jar his bone structure?
14301.6N
Step-by-step explanation:
What is the average acceleration of the man from the time his feet first touch the patio to the time he is brought fully to rest?
(Note that to solve this question you need to know and use the third equation of motion, v²=u²+2as, where v is final velocity, u is initial velocity, a is acceleration, and s is displacement.)
First the man drops 0.5m to the patio, and then it takes 2.9cm to fully stop. Let's look at the first half of this motion, from when he drops to when he first strikes the patio, but before he fully stops:
He drops to the patio, he doesn't jump with any momentum, so we can deduce his initial velocity (u) is 0m/s. The acceleration is due to gravity, so we take 'a' to be 9.8m/s/s, and the window is 0.5m above ground so s is 0.5. Subbing these in we get:
v²=u²+2as
v²=0²+2(9.8)(0.5)=9.8
v=3.13m/s, so the man strikes the patio at 3.13m/s
Now let's look at the part from when he first strikes the patio to when he fully comes to rest. He strikes the patio at 3.13m/s as we just figured out, so his initial velocity for this part is 3.13. We're told it takes 2.9cm to stop fully, so now s is 0.029. And if he's coming to a full rest, his final velocity will be 0. Subbing these in we get:
v²=u²+2as
0²=3.13²+2a(0.029)
0=9.8+0.058a
a=-9.8/0.085= -168.97m/s/s (value is neg because he comes to rest)
So the average acceleration is 168.97m/s/s
With what force does this jump jar his bone structure?
For this question we need to use Newton’s second law, F = ma + mg, where F is force, m is mass, a is acceleration and g is gravity:
F = ma + mg
F = m(a+g)
F = 80(168.97+9.8)=80(178.77)=14301.6
So the force exerted is 14301.6N