A lowly high diver pushes off horizontally with a speed of${}^{\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}$from the platform edge${}^{\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}}$ above the surface of the water. (a) At what horizontal distance from the edge is the diver${}^{\mathbf{0}\mathbf{.}\mathbf{800}\mathbf{}\mathit{s}}$ after pushing off? (b)At what vertical distance above the surface of the water is the diver just then? (c)At what horizontal distance from the edge does the diver strike the water?

Height of the platform${}^{{\mathrm{y}}_0=10.0\mathrm{m}}$

$$\begin{gathered} v_0=2.00m/s \\ \mathrm{a}=\mathrm{g}=-9.8\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The problem deals with the projectile motion of an object. The projectile motion of an object is the motion of an object thrown into the air, subject to the acceleration due to gravity. As the diver pushes himself horizontally so the initial vertical velocity for the diver is zero. The time and the velocity are known, so the horizontal and vertical distance can be found. When the diver stroked the water, it covered the distance of 10.0 m, the time required to cover that distance can be calculated. With the calculated time, the horizontal distance with which the diver will strike the water can be calculated.

Formulae

The displacement in projectile motion is written as,

$\mathrm{y}-{\mathrm{y}}_0={\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2$ (i)

Where${}^{{\mathrm{v}}_0}$is the initial velocity

The speed in general can be written as,

$\mathrm{v}=\frac{\mathrm{d}}{\mathrm{t}}$ (ii)

Here d can be taken as horizontal distance

Using equation (ii) the horizontal distance distance is,

$$\begin{gathered} {\mathrm{d}}_{\mathrm{x}}=2.00\times 0.8 \\ {\mathrm{d}}_{\mathrm{x}}=1.6\mathrm{m} \end{gathered}$$

As the diver pushes himself horizontally, the vertical velocity of the diver is zero (i.e.${\mathrm{v}}_{0\mathrm{y}}=0\frac{\mathrm{m}}{\mathrm{s}}$) thus the equation (i) can be written as

$$\begin{gathered} \mathrm{y}-10=0-\frac{1}{2}\times 9.8\times 0.8^2 \\ \mathrm{y}=10-3.136 \\ \mathrm{y}=6.86\mathrm{m} \end{gathered}$$

$$\begin{gathered} y-y_0=v_{0y}t+\frac{1}{2}\times a{t}^2 \\ -10=0-\frac{1}{2}\times 9.8\times t^2 \\ 10\times 2=9.8\times t^2 \\ \frac{20}{9.8}=t^2 \\ {t}^2=2.041 \\ t=\sqrt{2.041} \\ t=1.43s \end{gathered}$$

Again using equation (ii), the horizontal distance as the diver strikes the water is given by,

$$\begin{gathered} d_x=2.00\times 1.43 \\ {d}_{x}2.86\mathrm{m} \end{gathered}$$