3.10 A daring 510-N swimmer dives off a cliff with a running horizontal leap, as shown in Fig. E3.10. What must her minimum speed be just as she leaves the top of the cliff so that she will
miss the ledge at the bottom, which is 1.75 m wide and 9.00 m below the top of the cliff?
SOLUTION
The person moves in projectile motion once it takes off from the cliff. I will place my frame of reference at the edge of the cliff where –y-direction is downward.
Specifically, since the person will move half of a parabolic path thus at max height, thus the following are applicable,
$v_{0y}=0$
$v_{0x}= v_x$
$a_x=0$ for projectile motion
$a_y=-g$ (negative since my –y direction is downward)
And at the origin, $x_0 =0$ and $y_0 =0$
We are given with a height $y=-9.00 m$ (negative since my –y direction is downward) and a ledge whose distance is $x=1.75 m$ from the cliff.
To start you can solve either using variable x or y in the constant motion acceleration formula. Let us try using the x-variable first.
$x=x_0+v_{0x}t+\frac{1}{2}a_x t^2$
$ x=0+v_{0x}t+0$
$ v_{0x}=\frac{x}{t}\ \ \ $ (equation 1)
However, we can not solve this without time $t$, thus we use the given y-variable to solve for time.
$y=y_0+v_{0y}t+\frac{1}{2}a_yt^2$
$ y=0+0+\frac{1}{2}a_yt^2$
$t=\sqrt{\frac{2y}{-g}}$
$t=\sqrt{\frac{2(-9.00 m)}{-9.8m/s^2}}$
$t=1.36s$
Now, to solve for our initial velocity (equation 1) in order to miss the ledge x = 1.75 m below,
$ v_{0x}=\frac{1.75 \ m}{1.36 \ s }$
$ v_{0x}=1.29 \ \frac{m}{s}$
Thus, the swimmer must have an initial velocity of 1.29 m/s in order not to hit the ledge.