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:
[tex]v^2-u^2=2as[/tex]
where
v is the final velocity
u = 0 is the initial velocity (the diver starts from rest)
[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity
s is the displacement
For the diver jumping from 5 m, s = 5 m, so
[tex]v=\sqrt{2as}=\sqrt{2(9.8)(5)}=9.9 m/s[/tex]
For the diver jumping from 10 m, s = 10 m, so
[tex]v=\sqrt{2as}=\sqrt{2(9.8)(10)}=14 m/s[/tex]
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
[tex]s=ut+\frac{1}{2}at^2[/tex]
And since u = 0, it can be reduced to
[tex]s=\frac{1}{2}at^2[/tex]
For the diver jumping from 5 m, s = 5 m, so we find
[tex]t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(5)}{9.8}}=1.01 s[/tex]
For the diver jumping from 10 m, s = 10 m, so we find
[tex]t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(10)}{9.8}}=1.43 s[/tex]
