A person jumps from a window 57 meters high and is caught in a firefighter's net which stretches 0.6 meter. To the nearest tenth of a m/s^2, what is the magnitude of the person's acceleration in the net?
mgh=f*d=mad
solve for a
Bobpursley, if you could, would be able to elaborate on what you mean by "mgh=f*d=mad"
I am new to physics and I have pretty much this same exact problem for my homework and I can't figure it out on my own. I have searched the book and nothing that is in the book that has been covered thus far seems to click as for how I can solve this problem.
To find the magnitude of the person's acceleration in the net, we can use the following equation of motion:
v^2 = u^2 + 2 * a * s
Where:
v = final velocity (0 m/s, as the person comes to a stop)
u = initial velocity (unknown)
a = acceleration (what we need to find)
s = displacement (57 m + 0.6 m = 57.6 m)
Since the person jumps from a window, there is no initial vertical velocity (u = 0 m/s), so we can simplify the equation to:
v^2 = 2 * a * s
Substituting the given values:
(0 m/s)^2 = 2 * a * 57.6 m
0 = 2 * a * 57.6 m
Now, let's solve for a:
2 * a * 57.6 m = 0
a * 57.6 m = 0
a = 0 / 57.6 m
a = 0 m/s^2
Therefore, the magnitude of the person's acceleration in the net is 0 m/s^2.