Question

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.

Answer 1

Answer: 23000 frames/s

Step-by-step 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 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 (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=(d_(max))/(V_(i))`

`t=(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=(1)/(t)`

`f=(1)/(0.00004347 s/frame)`

Finally:

`f=23000 frames/s`

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