A daring 510 Nswimmer 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 is1.75 mwide and9.00 mbelow the top of the cliff.

The given data can be listed below,

The height of the cliff is $\mathrm{h}=9\mathrm{m}$.

The width of the ledge is $\mathrm{x}=1.75\mathrm{m}$.

Acceleration due to gravity, $\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2$

When a moving body has a uniform velocity, its speed changes constantly with regard to time, or gains in speed are directly proportionate to longer periods of time.

The trajectory of the person is given by,

As the initial vertical speed of the person is zero. the time taken by the person to reach the cliff is given by,

$$\begin{gathered} \mathrm{h}={\mathrm{v}}_{0\mathrm{y}}+\frac{1}{2}{\mathrm{gt}}^2 \\ \mathrm{t}=\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}} \end{gathered}$$

Here, h is the height of the cliff and g is the acceleration due to gravity.

Substitute all the values in the above,

$$\begin{gathered} \mathrm{t}=\sqrt{\frac{2\times 9\mathrm{m}}{9.8\mathrm{m}/{\mathrm{s}}^2}} \\ =1.36\mathrm{s} \end{gathered}$$

The minimum speed of the person to leave the cliff is given by,

$$\begin{gathered} \mathrm{x}={\mathrm{v}}_{0\mathrm{x}}\mathrm{t} \\ {\mathrm{v}}_{0\mathrm{x}}=\frac{\mathrm{x}}{\mathrm{t}} \end{gathered}$$

Here, x is the horizontal width of the ledge and t is the time taken to reach the cliff.

Substitute all the values in the above,

$$\begin{gathered} v_{0x}=\frac{1.75\mathrm{m}}{1.36\mathrm{s}} \\ =1.29\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the minimum speed of the person to leave the cliff is 1.29m/s .