Question

Trained dolphins are capable of a vertical leap of 7.0m straight up from the surface of the water-an impressive feat.Suppose you could train a dolphin to launch itself out of the water at this same speed but at an angle.
What maximum horizontal range could the dolphin achieve?

Answer 1

Answer: 14 m

Explanation:

This situation is related to parabolic motion, in which the motion of the dolphins has two components: x-component and y-component. Being their main equations as follows:  

x-component:  

$x=V_{ox} t$   (1)  

$V_{ox}=V_{o}cos\theta$ (2)  

Where:  

$V_{o}$ is the dolphin's initial speed  

$\theta$ is the angle

$t$ is the time since the dolphin jumps until it goes to the water surface again

$V_{ox}$ is the horizontal component of the initial velocity

y-component:  

$y_{max}=\frac{V_{o}^{2} (sin \theta)^{2}}{2 g}$ (3)

$V_{y}=V_{oy}-gt$   (4)  

$V_{oy}=V_{o}sin\theta$   (5)  

Where:  

$y_{max}=7 m$ is the maximum height the dolphin can reach (when $V_{y}=0$)  

$V_{y}=0$ is the velocity of the dolphin at its maximum height

$g=9.8 m/s^{2}$ is the acceleration due gravity

$V_{oy}$ is the vertical component of the initial velocity

Let's begin with (3) when $\theta=90\°$ (straight up leap):

$7 m=\frac{V_{o}^{2}}{2(9.8 m/s^{2})}$ (6)

Finding $V_{o}$:

$V_{o}=11.71 m/s$ (7)

Isolating $t$ from (4):

$t=\frac{V_{oy}}{g}=\frac{V_{o} sin\theta}{g}$   (8)  

Substituting (7) and (8) in (1):

$x=V_{o}cos\theta(\frac{V_{o} sin\theta}{g})$ (9)  

$x=\frac{V_{o}^{2} sin(2\theta)}{g}$ (10)  

$x$ is maximum when $sin(2\theta)=1$. This means:

$2\theta=sin^{-1} 1$

$\theta=\frac{90\°}{2}$

$\theta=45\°$

Hence, the maximum horizontal range the dolphin could achieve is when $\theta=45\°$:

$x=\frac{(11.71 m/s)^{2} sin(2(45\°))}{9.8 m/s^{2}}$ (10)  

Finally:

$x=13.99 m \approx 14 m$