The swimmer dives only with horizontal speed.

Therefore, v_{0}_{x} = v_{0}

v_{0}_{y} = 0.

Let's consider the kinematic equation for the motion in the y-direction

$\overline{){\mathbf{\u2206}}{\mathbf{y}}{\mathbf{=}}{{\mathbf{v}}}_{{\mathbf{0}}{\mathbf{y}}}{\mathbf{t}}{\mathbf{-}}\frac{\mathbf{1}}{\mathbf{2}}{{\mathbf{gt}}}^{{\mathbf{2}}}}$

For the x-direction, we'll use:

$\overline{){\mathbf{\u2206}}{\mathbf{x}}{\mathbf{=}}{{\mathbf{v}}}_{{\mathbf{0}}{\mathbf{x}}}{\mathbf{t}}{\mathbf{+}}\frac{\mathbf{1}}{\mathbf{2}}{{\mathbf{at}}}^{{\mathbf{2}}}}$

To get the time, t, we'll use the equation in the y-direction.

We know that v_{0y} = 0

A daring 510-N swimmer dives off a cliff with a running horizontal leap, as shown in the figure .

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?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Projectile Motion: Horizontal & Negative Launch concept. If you need more Projectile Motion: Horizontal & Negative Launch practice, you can also practice Projectile Motion: Horizontal & Negative Launch practice problems.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Olson's class at DAYTONA STATE.