Question

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.

Answer 1

Answer: 23000 frames/s

Explanation:

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

$V_{i}=4,7 m/s$

In addition, we know the maximum distance 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.