1) Velocity: 9.9 m/s and 14 m/s
The motion of the diver is a free-fall motion, so it is a uniform accelerated motion. Choosing downward as positive direction, the final velocity can be found by using the following suvat equation:
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
where
v is the final velocity
u = 0 is the initial velocity (the diver starts from rest)
is the acceleration of gravity
s is the displacement
For the diver jumping from 5 m, s = 5 m, so
![v=\sqrt{2as}=\sqrt{2(9.8)(5)}=9.9 m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B2as%7D%3D%5Csqrt%7B2%289.8%29%285%29%7D%3D9.9%20m%2Fs)
For the diver jumping from 10 m, s = 10 m, so
![v=\sqrt{2as}=\sqrt{2(9.8)(10)}=14 m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B2as%7D%3D%5Csqrt%7B2%289.8%29%2810%29%7D%3D14%20m%2Fs)
2) Time: 1.01 s and 1.43 s
The time of flight of each diver can be found by using the other suvat equation
![s=ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
And since u = 0, it can be reduced to
![s=\frac{1}{2}at^2](https://tex.z-dn.net/?f=s%3D%5Cfrac%7B1%7D%7B2%7Dat%5E2)
For the diver jumping from 5 m, s = 5 m, so we find
![t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(5)}{9.8}}=1.01 s](https://tex.z-dn.net/?f=t%3D%5Csqrt%7B%5Cfrac%7B2s%7D%7Ba%7D%7D%3D%5Csqrt%7B%5Cfrac%7B2%285%29%7D%7B9.8%7D%7D%3D1.01%20s)
For the diver jumping from 10 m, s = 10 m, so we find
![t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(10)}{9.8}}=1.43 s](https://tex.z-dn.net/?f=t%3D%5Csqrt%7B%5Cfrac%7B2s%7D%7Ba%7D%7D%3D%5Csqrt%7B%5Cfrac%7B2%2810%29%7D%7B9.8%7D%7D%3D1.43%20s)