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: Â
x-component: Â
[tex]x=V_{ox} t[/tex] Â (1) Â
[tex]V_{ox}=V_{o}cos\theta[/tex] (2) Â
Where: Â
[tex]V_{o}[/tex] is the dolphin's initial speed Â
[tex]\theta[/tex] is the angle
[tex]t[/tex] is the time since the dolphin jumps until it goes to the water surface again
[tex]V_{ox}[/tex] is the horizontal component of the initial velocity
y-component: Â
[tex]y_{max}=\frac{V_{o}^{2} (sin \theta)^{2}}{2 g}[/tex] (3)
[tex]V_{y}=V_{oy}-gt[/tex] Â (4) Â
[tex]V_{oy}=V_{o}sin\theta[/tex] Â (5) Â
Where: Â
[tex]y_{max}=7 m[/tex] is the maximum height the dolphin can reach (when [tex]V_{y}=0[/tex]) Â
[tex]V_{y}=0[/tex] is the velocity of the dolphin at its maximum height
[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity
[tex]V_{oy}[/tex] is the vertical component of the initial velocity
Let's begin with (3) when [tex]\theta=90\°[/tex] (straight up leap):
[tex]7 m=\frac{V_{o}^{2}}{2(9.8 m/s^{2})}[/tex] (6)
Finding [tex]V_{o}[/tex]:
[tex]V_{o}=11.71 m/s[/tex] (7)
Isolating [tex]t[/tex] from (4):
[tex]t=\frac{V_{oy}}{g}=\frac{V_{o} sin\theta}{g}[/tex] Â (8) Â
Substituting (7) and (8) in (1):
[tex]x=V_{o}cos\theta(\frac{V_{o} sin\theta}{g})[/tex] (9) Â
[tex]x=\frac{V_{o}^{2} sin(2\theta)}{g}[/tex] (10) Â
[tex]x[/tex] is maximum when [tex]sin(2\theta)=1[/tex]. This means:
[tex]2\theta=sin^{-1} 1[/tex]
[tex]\theta=\frac{90\°}{2}[/tex]
[tex]\theta=45\°[/tex]
Hence, the maximum horizontal range the dolphin could achieve is when [tex]\theta=45\°[/tex]:
[tex]x=\frac{(11.71 m/s)^{2} sin(2(45\°))}{9.8 m/s^{2}}[/tex] (10) Â
Finally:
[tex]x=13.99 m \approx 14 m[/tex]
