To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,
![m= 89 kg\\x = 3.1 m\\t = 0.5s\\a = g = 9.8m/s^2](https://tex.z-dn.net/?f=m%3D%2089%20kg%5C%5Cx%20%3D%203.1%20m%5C%5Ct%20%3D%200.5s%5C%5Ca%20%3D%20g%20%3D%209.8m%2Fs%5E2)
Through the aforementioned formula we will have to
![v_f^2-v_i^2 = 2ax](https://tex.z-dn.net/?f=v_f%5E2-v_i%5E2%20%3D%202ax)
The particulate part of the rest, so the final speed would be
![v_f^2 = 2gx](https://tex.z-dn.net/?f=v_f%5E2%20%3D%202gx)
![v_f=\sqrt{2(9.8)(3.1)}](https://tex.z-dn.net/?f=v_f%3D%5Csqrt%7B2%289.8%29%283.1%29%7D)
![v_f = 7.79m/s](https://tex.z-dn.net/?f=v_f%20%3D%207.79m%2Fs)
Now from Newton's second law we know that
![F = ma](https://tex.z-dn.net/?f=F%20%3D%20ma)
Here,
m = mass
a = acceleration, which can also be written as a function of velocity and time, then
![F = m\frac{dv}{dt}](https://tex.z-dn.net/?f=F%20%3D%20m%5Cfrac%7Bdv%7D%7Bdt%7D)
Replacing we have that,
![F = (89)\frac{7.79}{0.5}](https://tex.z-dn.net/?f=F%20%3D%20%2889%29%5Cfrac%7B7.79%7D%7B0.5%7D)
![F = 1386.62N](https://tex.z-dn.net/?f=F%20%3D%201386.62N)
Therefore the force that the water exert on the man is 1386.62
The resultant vector can be determined by the component vectors. The component vectors are vector lying along the x and y-axes. The equation for the resultant vector, v is:
v = √(vx² + vy²)
v = √[(9.80)² + (-6.40)²]
v = √137 or 11.7 units
Some examples of constant velocity (or at least almost- constant velocity) motion include (among many others): • A car traveling at constant speed without changing direction. A hockey puck sliding across ice. A space probe that is drifting through interstellar space.
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