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Helen [10]
3 years ago
12

A dolphin in an aquatic show jumps straight up out of the water at a velocity of 11.8 m/s. The magnitude of the gravitational ac

celeration g = 9.8 m/s2 Take the water surface to be y0 = 0. Choose UPWARD as positive y direction. Keep 2 decimal places in all answers. (a) How high (what maximum height) in meters does his body rise above the water? To solve this part, first note that the body's final velocity at the maximum height is known (implicitly) and identify its value. Then identify the unknown, and chose the appropriate equation (based on the knowns and unknown) to solve for it. After choosing the equation, solving for the unknown, checking units, think about whether the answer is reasonable.(b) How long in seconds is the dolphin in the air? Neglect any effects due to his size or orientation. Note: this is the total time of jumping up from water to hightest point and falling down to water.
Physics
1 answer:
siniylev [52]3 years ago
4 0

Answer:

a) y_{max}=7.10m

b) t=2.40 s

Explanation:

From the exercise, we know the <u>initial velocity</u><u>, </u><u>gravitational acceleration and initial position of the dolphin</u><u>.</u>

v_{oy}=11.8m/s

y_{o}=0m\\ g=9.8m/s^{2}

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 <u>time</u> does it take the dolphin to reach maximum height.

v_{y}=v_{oy}+gt

0=11.8m/s-(9.8m/s^{2} )t

Solving for t

t=1.20 s

So, the dolphin reach maximum point at 1.20 seconds

Now, using the equation of position we can calculate <u>maximum height</u>.

y=y_{o}+v_{oy}t +\frac{1}{2}gt^{2}

y=0+11.8m/s(1.20s)-\frac{1}{2}(9.8m/s^{2} )(1.20s)=7.10m

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

y=y_{o}+v_{oy}t +\frac{1}{2}gt^{2}

0=0+11.8t-\frac{1}{2}(9.8)t^{2}

What we have here, is a quadratic equation that could be solve using:

t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}

a=-\frac{1}{2} (9.8)

b=11.8\\c=0

t=0s or t=2.40 s

Since the answer can not be 0, the <u>dolphin is 2.40 seconds in the air</u>.

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