The height of the swimmer's feet in the air at time [tex]t[/tex] is given according to
[tex]y=1.20\,\mathrm m+\left(4.00\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2[/tex]
where [tex]g[/tex] is the magnitude of the acceleration due to gravity (taken here to be 9.80 m/s^2).
a. Solve for [tex]t[/tex] when [tex]y=0[/tex]:
[tex]1.20\,\mathrm m+\left(4.00\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies\boxed{t=1.05\,\mathrm s}[/tex]
(The other solution is negative; ignore it)
b. At her highest point [tex]y_{\rm max}[/tex], the swimmer has zero velocity, so
[tex]-\left(4.00\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1.20\,\mathrm m)\implies\boxed{y_{\rm max}=2.02\,\mathrm m}[/tex]
c. Her velocity at time [tex]t[/tex] is
[tex]v=4.00\dfrac{\rm m}{\rm s}-gt[/tex]
After 1.05 s in the air, her velocity will be
[tex]v=4.00\dfrac{\rm m}{\rm s}-g(1.05\,\mathrm s)\implies\boxed{v=-6.29\dfrac{\rm m}{\rm s}}[/tex]
