A 30-ft tall drop tower is being used to study the shape of water droplets as they fall through air. A camera is to be carried by a cam-operated linkage which will track the droplets motion from the 7-ft to the 9-ft point in its fall (measured from release point at the top of the tower). The drops are released every 0.5 sec. Every drop is to be filmed. Design a cam and follower which will track these droplets, matching their velocities and accelerations in the 1-ft filming window. Consider acceleration due to gravity, g = 32.2 ft/sec2 . Note that the follower is operating between 7-ft to the 9-ft but it follows the water droplets exactly within the 1-ft filming window.

The camera would flow the same velocity and acceleration of the droplet until 9ft. After that it would switch gear and accelerate back to its original spot at the rate of 140.54 m/s2 in order to catch the next droplet.

Explanation:

The time it takes for the drop to reach 7-ft and 9-ft at the acceleration of 32.2ft/s2:

When s = 7ft,

When s = 9ft,

So it would take Δt = 0.75 - 0.66 = 0.09 s for each drop to travel from 7ft to 9ft. As the drops are released every 0.5s, this means that after our cam follow the first drop to the end of the 7-9ft window, it has 0.5 - 0.09 = 0.41 s or less to reverse its acceleration and go back to the original spot to take care of the next drop.

So at 9ft, the drop and camera velocity would be:

So if the camera reverses its direction to go back to original spot at the rate of a (ft/s2) and initial speed of 24.1 ft/s. Within the time span of 0.41s to catch the next drop at 7ft

So the camera would flow the same velocity and acceleration of the droplet until 9ft. After that it would switch gear and accelerate back to its original spot at the rate of 140.54 m/s2 in order to catch the next droplet.

answer:

okay

explanation:

yes indeed

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