# 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  $ax^2 + bx +c = 0.$

#### First Step

Choose the origin at the diving board.

Be careful of signs.

Consider only motion in the $y$ direction.

#### Calculations

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

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

$t = 1.85\; s$