(a) Time of flight: 1.01 s and 1.43 s
The motion of a diver jumping from a platform and entering the water is a free-fall motion, so its vertical displacement is given by
[tex]s=ut+\frac{1}{2}gt^2[/tex]
where
u = 0 is the initial velocity
t is the time
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity (taking downward as positive direction)
If we re-arrange the equation, we can solve for t, the time it takes for each diver to enter the water:
[tex]t=\sqrt{\frac{2s}{g}}[/tex]
For the diver jumping from 5 m, s = 5 m, so we get
[tex]t=\sqrt{\frac{2(5)}{9.8}}=1.01 s[/tex]
For the diver jumping from 10 m, s = 10 m, so we get
[tex]t=\sqrt{\frac{2(10)}{9.8}}=1.43 s[/tex]
(b) final velocity: 9.9 m/s and 14.0 m/s
In order to find the final velocity of each diver as they enter the water, we can now use the following suvat equation:
[tex]v=u+gt[/tex]
where
v is the final velocity
u = 0 is the initial velocity
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
t is the time at which the diver enters the water
For the diver jumping from 5 m, t = 1.01 s, so the final velocity is
[tex]v=0+(9.8)(1.01)=9.9 m/s[/tex]
For the diver jumping from 10 m, t = 1.43 s, so the final velocity is
[tex]v=0+(9.8)(1.43)=14.0 m/s[/tex]
