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:

c. 25,000 frames/s

Explanation:

For computing the minimum frame rate for high speed first we have to determine the time by applying the following equation

$t = \frac{d}{s}$

$= \frac{0.2\ mm}{4.6\ m/s }$

$= \frac{0.2 \times 10 ^{-3}}{4.6\ m/s }$

$= 4.347 \times 10^{-5} sec$

Now the frame rate is

$Frame\ rate = \frac{1}{t}$

$= \frac{1}{4.347 \times 10^{-5} sec}$

= 23,000 frame per sec

≈ 25,000 frame per sec

First we have find the time then after finding out the time we calculate the frame time by applying the above formula so that the minimum frame rate could come