SOLUTION
Place your frame of reference at the edge of the cliff where the -y-direction points downward.
The moment the swimmer takes the leap it becomes a projectile. Specifically, it is half of a parabolic path thus at max height the following applies:
$x=1.75m$ and $y=9.00m$
$v_{0y}=0$
$a_x = 0$
Using the equation, $v_x=v_{0x}+a_xt=v_{0x}+0$
$v_x=v_{0x}$
$a_x=0$ and $a_y=-g$
At the point of origin $x_0=y_0=0$
Since -y direction is pointed below then $y=-9.00m$ and $x=1.75 \ m$ then you may solve using the equation $x=x_0+v_{0x}t+\frac{1}{2}a_xt^2$ or $y=y_0+v_{0y}t+\frac{1}{2}a_yt^2$.
Let us start with using the x-variable.
$x=x_0+v_{0x}t+\frac{1}{2}a_xt^2$
$x=0+v_{0x}t+0$
so
$v_{0x}=v_x=\frac{x}{t}$ [equation 1]
Let us find for time t, using
$y=y_0+v_{0y}t+\frac{1}{2}a_yt^2$
$y=0+0+\frac{1}{2}a_yt^2$
$t=\sqrt{\frac{2y}{-g}}$
$t=\sqrt{\frac{2(-9.00m)}{-9.8m/s^2}}$
$t= 1.36 s$
Substitute our time t=1.36 seconds to equation 1
$v_{0x}=\frac{x}{t}$
$v_{0x}=\frac{1.75m}{1.36s}$
$v_{0x}=v_x=1.29 \ m/s$
Thus, the swimmer must have a minimum speed of at least 1.29 m/s in order not to die from crashing on the ledge.
Reference: Young, H. D. et al. 2012. Sear’s and Zemansky’s University Physics with Modern Physics. NY: Pearson Education Inc.
