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Karolina [17]
3 years ago
15

Some plants disperse their seeds by having the fruit split and contract, propelling the seeds through the air. The trajectory of

these seeds can be determined by using a high-speed camera. In one type of plant, seeds are projected at 20 cm above ground level with initial speeds between 2.3 m/s and 4.7 m/s . The launch angle is measured from the horizontal, with +90∘ corresponding to an initial velocity straight upward and −90∘ straight downward.
Physics
1 answer:
Pani-rosa [81]3 years ago
3 0

Answer: 23000 frames/s

Explanation:

The rest of the statement of the question is presented below:

The experiment is designed so that the seeds move no more than 0.20 mm between photographic frames. What minimum frame rate for the high-speed camera is needed to achieve this?

We know the <u>maximum initial speed</u> at which the seeds are dispersed is:

V_{i}=4,7 m/s

In addition, we know the <u>maximum distance</u> at which the seeds move between photographic frames is:

d_{max}=0.20 mm \frac{1m}{1000 m}=0.0002 m

And we need to find the minimum frame rate of the camera with these given conditions. This can be found by finding the time t for each frame and then the frame rate:

Finding the time:

t=\frac{d_{max}}{V_{i}}

t=\frac{0.0002 m}{4.6 m/s}

t=0.00004347 s/frame This  is the time for each frame

Now we need to find the frame rate, which is the frequency at which the photos are taken.

In this sense, frequency f is defined as:

f=\frac{1}{t}

f=\frac{1}{0.00004347 s/frame}

Finally:

f=23000 frames/s

Hence, the minimum frame rate is 23000 frames per second.

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