Question

A diver leaves the end of a 4.0-m-high diving board and strikes the water 1.3 s later, 3.0 m beyond the end of the board. Considering the diver as a particle, determine: (a) her initial velocity,${\overrightarrow{v}}_0$, (b) the maximum height reached; and (c) the velocity${\overrightarrow{v}}_{\mathrm{f}}$with which she enters the water.

Short answer

(a) The initial velocity vector of the diver has a magnitude of $4.02\mathrm{m}/\mathrm{s}$that makes an angle of $55°$with the horizontal.

(b) The maximum height reached by the diver is 4.6 m.

(c) The diver enters the water with a final velocity of $9.7\mathrm{m}/\mathrm{s}$, thereby making an angle of $76.3°$with the positive horizontal in the clockwise direction.

Step-by-step solution

Step 1. Kinematic equations for projectile motion

When a particle is projected from a height yabove the ground with an initial velocity${\overrightarrow{v}}_0$, it falls downward, following a parabolic path. The particle that is projected is known as the projectile.

If the point of projection is taken as the origin and the upward direction as the positive y-axis, then the horizontal and vertical components of acceleration will be: $a_{\mathrm{x}}=0$and $a_{\mathrm{y}}=-g$Thus, the kinematics equations for projectile motion are as follows:

• For the horizontal motion of projectile:

$\begin{array}{c}{v}_{\mathrm{x}}=v_{\mathrm{x}0}\\ x=v_{\mathrm{x}0}t\end{array}$

• For the vertical motion of projectile:

$\begin{array}{c}{v}_{\mathrm{y}}=v_{\mathrm{y}0}-gt\\ y=v_{\mathrm{y}0}t-\frac{1}{2}g{t}^2\\ v_{\mathrm{y}}^2=v_{\mathrm{y}0}^2-2gy\end{array}$

Here, xandy are the horizontal and vertical displacements of the projectile traveled in time t.

$v_{\mathrm{x}0}$and $v_{\mathrm{y}0}$are the horizontal and vertical components of the initial velocity of the particle.

$v_{\mathrm{x}},v_{\mathrm{y}}$are the horizontal and vertical components of the velocity of the particle at time t.

Step 2. Given information

Here, the diver is the projectile. Take the endpoint of the diving board as the origin and the upward direction as the positive y-axis. This is shown in the diagram below:

Thus, the height of the diving board is $y=-4.0\mathrm{m}$.

The horizontal distance traveled by the particle is $x=3.0\mathrm{m}$.

The time taken by the particle to reach the surface of the water is $t=1.3\mathrm{s}$.

Step 3. (a) Determination of the initial velocity of the diver (particle)

Let $v_{\mathrm{x}0}$and $v_{\mathrm{y}0}$be the horizontal and vertical components of the initial velocity ${\overrightarrow{v}}_0$of the particle.

Using the kinematic equation for the horizontal motion of the projectile, you will get $v_{\mathrm{x}0}$as:

$\begin{array}{c}x=v_{\mathrm{x}0}t\\ v_{\mathrm{x}0}=\frac{x}{t}\\ =\frac{\left(3.0\mathrm{m}\right)}{\left(1.3\mathrm{s}\right)}\\ =2.31\mathrm{m}/\mathrm{s}\end{array}$

Using the kinematic equation for the vertical motion of the projectile, you will get $v_{\mathrm{y}0}$as:

$\begin{array}{c}y=v_{\mathrm{y}0}t-\frac{1}{2}g{t}^2\\ \left(-4.0\mathrm{m}\right)=v_{\mathrm{y}0}\left(1.3\mathrm{s}\right)-\frac{1}{2}\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(1.3\mathrm{s}\right)}^2\\ v_{\mathrm{y}0}=\frac{\left(-4.0\mathrm{m}\right)}{\left(1.3\mathrm{s}\right)}+\frac{1}{2}\frac{\left(9.8\mathrm{m}/{\mathrm{s}}^2\right){\left(1.3\mathrm{s}\right)}^2}{\left(1.3\mathrm{s}\right)}\\ =-3.08\mathrm{m}/\mathrm{s}+6.37\mathrm{m}/\mathrm{s}\\ =3.29\mathrm{m}/\mathrm{s}\end{array}$

Thus, the magnitude of the initial velocity of the particle is:

$\begin{array}{c}{v}_0=\sqrt{v_{\mathrm{x}0}^2+v_{\mathrm{y}0}^2}\\ =\sqrt{{\left(2.31\mathrm{m}/\mathrm{s}\right)}^2+{\left(3.29\mathrm{m}/\mathrm{s}\right)}^2}\\ =\sqrt{16.16{\mathrm{m}}^2/{\mathrm{s}}^2}\\ =4.02\mathrm{m}/\mathrm{s}\end{array}$

If $\theta$ is the angle which the initial velocity vector makes with the horizontal, then:

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{v_{\mathrm{y}0}}{v_{\mathrm{x}0}}\right)\\ ={\tan }^{-1}\left(\frac{3.29}{2.31}\right)\\ ={\tan }^{-1}\left(1.424\right)\\ =54.9°\approx 55°\end{array}$

Thus, theinitial velocity of the particle is$4.02\mathrm{m}/\mathrm{s}$and it makes an angle of$55°$with the horizontal.

(b) Determination of the maximum height reached by the diver (particle)

At the point of the maximum height, the velocity of the particle will be along the horizontal; therefore, the y component of velocity will be zero, i.e.,$v{\text{'}}_{\mathrm{y}}=0\mathrm{m}/\mathrm{s}$. Let the height of this point above the reference level be $y\text{'}$.

Using the kinematic equation for the vertical motion of the projectile at this point, you will get:

$\begin{array}{c}{\left({v\text{'}}_y\right)}^2=v_{\mathrm{y}0}^2-2\mathrm{g}y\text{'}\\ y\text{'}=\frac{v_{\mathrm{y}0}^2-{\left({v\text{'}}_{\mathrm{y}}\right)}^2}{2\mathrm{g}}\\ =\frac{{\left(3.29\mathrm{m}/\mathrm{s}\right)}^2-{\left(0\mathrm{m}/\mathrm{s}\right)}^2}{2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}\\ =0.6\mathrm{m}\end{array}$

Thus, the maximum height reached by the particle is:

$\begin{array}{c}y+y\text{'}=\left(4+0.6\right)\mathrm{m}\\ =4.6\mathrm{m}\end{array}$

Step 5. (c) Determination of the final velocity of the diver with which she enters the surface of the water

Let $v_{\mathrm{x}},v_{\mathrm{y}}$ be the horizontal and vertical components of the final velocity$v_{\mathrm{f}}$ of the particle at time twhen it enters the surface of the water.

Using the kinematic equation for the vertical motion of the projectile, you will get $v_{\mathrm{y}}$as:

$\begin{array}{c}{v}_y=v_{y0}-\mathrm{g}t\\ =\left(3.29\mathrm{m}/\mathrm{s}\right)-\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(1.3\mathrm{s}\right)\\ =-9.45\mathrm{m}/\mathrm{s}\end{array}$

Since the horizontal velocity of the particle remains the same, the horizontal component of the final velocity will be:

$v_{\mathrm{x}}=v_{\mathrm{x}0}=2.31\mathrm{m}/\mathrm{s}$

The magnitude of the final velocity of the particle is:

$\begin{array}{c}{v}_{\mathrm{f}}=\sqrt{v_{\mathrm{x}}^2+v_{\mathrm{y}}^2}\\ =\sqrt{{\left(2.31\mathrm{m}/\mathrm{s}\right)}^2+{\left(-9.45\mathrm{m}/\mathrm{s}\right)}^2}\\ =\sqrt{94.64{\mathrm{m}}^2/{\mathrm{s}}^2}\\ =9.7\mathrm{m}/\mathrm{s}\end{array}$

Thus, the magnitude of the final velocity of the particle is $9.7\mathrm{m}/\mathrm{s}$.

If$\theta \text{'}$ is the angle which the initial velocity vector makes with the horizontal, then:

$\begin{array}{c}\theta \text{'}={\tan }^{-1}\frac{v_{\mathrm{y}}}{v_{\mathrm{x}}}\\ ={\tan }^{-1}\left(\frac{-9.45}{2.31}\right)\\ ={\tan }^{-1}\left(-4.09\right)\\ =-76.3°\end{array}$

Here, the negative sign shows that the final velocity of the particle lies under the surface of the water, thereby making an angle of $76.3°$with the positive horizontal in the clockwise direction.