A person jumps from a window 34 meters high and is caught in a firefighter's net which stretches 0.6 meter. To the nearest m/s^2, what is the magnitude of the person's acceleration in the net?
v=sqrt(2gH) = sqrt(2•9.8•34) = 25.8 m/s
the decelerated motion
a= [(v(fin) ² -v(init)²]/2•s=(0- 25.8²)/2•0.6=- 555.33≈ - 555 m/s²
To find the magnitude of the person's acceleration in the net, we can use the equation of motion for accelerated motion with constant acceleration:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, since the person comes to rest)
u = initial velocity (unknown)
a = acceleration (unknown)
s = displacement (34 m + 0.6 m)
First, let's find the initial velocity of the person. Since the person jumps, the initial velocity can be taken as 0 m/s.
v^2 = u^2 + 2as
0^2 = (0 m/s)^2 + 2a(34 m + 0.6 m)
0 = 2a(34.6 m)
Now we can solve for the acceleration (a):
0 = 69.2a
To find the magnitude of the acceleration, we disregard the negative sign. Therefore:
69.2a = 0
a = 0 m/s^2
Therefore, to the nearest m/s^2, the magnitude of the person's acceleration in the net is 0 m/s^2.