# Biophysics Problem 16

The great swimmer, Mark McSpitswater, dives from a diving board \(\mathrm{15.0\; m}\) above the water's surface. His initial velocity is \(2.00\; m/s^{-1}\) at an angle of \(30.0^\circ\) up from the horizontal.

(a) How long does it take him to hit the water? (Recall that \(x = (-b \pm \sqrt{b^2 - 4ac} )/2a\) is the solution to \(ax^2 + bx +c = 0.\)

Choose the origin at the diving board.

Be careful of signs.

Consider only motion in the \(y\) direction.

\(v_{0y} = 2 \sin30 = 1m/s \\ y = v_{0y}t + (1/2)a_yt^2\)

Choose origin at diving board then \(a = -9.8\; m/s^2, \;y = -15\;m\)

\(-15 = 1t - (1/2)(9.8)t^2 \\ t^2 -0.2041t - 3.061 = 0\)

Using the solution to the quadratic equation:

\(t = 0.10205 +/- 1.7525\)

Only the + solution is valid

\(t = 1.85\; s\)