Some plants disperse their seeds when the fruit splits and contracts, propelling the seeds through the air. The trajectory of these seeds can be determined with a high-speed camera. In an experiment on one type of plant, seeds are projected at 20 cm above ground level with initial speeds between 2.3 m/s and 4.6 m/s. The launch angle is measured from the horizontal, with +90∘ corresponding to an initial velocity straight up and −90∘ straight down. 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? (a) 250 frames/s; (b) 2500 frames/s; (c) 25,000 frames/s; (d) 250,000 frames/s.

Respuesta :

Answer: 23000 frames/s

Explanation:

We know the maximum initial speed at which the seeds are dispersed is:

[tex]V_{i}=4,7 m/s[/tex]

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

[tex]d_{max}=0.20 mm \frac{1m}{1000 m}=0.0002 m[/tex]

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

Finding the time:

[tex]t=\frac{d_{max}}{V_{i}}[/tex]

[tex]t=\frac{0.0002 m}{4.6 m/s}[/tex]

[tex]t=0.00004347 s/frame[/tex] 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 [tex]f[/tex] is defined as:

[tex]f=\frac{1}{t}[/tex]

[tex]f=\frac{1}{0.00004347 s/frame}[/tex]

Finally:

[tex]f=23000 frames/s[/tex]

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

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