Answer:
a) [tex]y_{max}=7.10m[/tex]
b) [tex]t=2.40 s[/tex]
Explanation:
From the exercise, we know the initial velocity, gravitational acceleration and initial position of the dolphin.
[tex]v_{oy}=11.8m/s[/tex]
[tex]y_{o}=0m\\ g=9.8m/s^{2}[/tex]
a) To find maximum height, we know that at that point the dolphin's velocity is 0 and it becomes coming down later.
Knowing that, we need to know how much time does it take the dolphin to reach maximum height.
[tex]v_{y}=v_{oy}+gt[/tex]
[tex]0=11.8m/s-(9.8m/s^{2} )t[/tex]
Solving for t
[tex]t=1.20 s[/tex]
So, the dolphin reach maximum point at 1.20 seconds
Now, using the equation of position we can calculate maximum height.
[tex]y=y_{o}+v_{oy}t +\frac{1}{2}gt^{2}[/tex]
[tex]y=0+11.8m/s(1.20s)-\frac{1}{2}(9.8m/s^{2} )(1.20s)=7.10m[/tex]
b) To find how long is the dolphin in the air we need to analyze it's hole motion
At the end of the jump the dolphin return to the water at y=0. So, from the equation of position we have that
[tex]y=y_{o}+v_{oy}t +\frac{1}{2}gt^{2}[/tex]
[tex]0=0+11.8t-\frac{1}{2}(9.8)t^{2}[/tex]
What we have here, is a quadratic equation that could be solve using:
[tex]t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}[/tex]
[tex]a=-\frac{1}{2} (9.8)[/tex]
[tex]b=11.8\\c=0[/tex]
[tex]t=0s[/tex] or [tex]t=2.40 s[/tex]
Since the answer can not be 0, the dolphin is 2.40 seconds in the air.
