Set 4 Problem number 20

Problem

Problem: An object initially moving upward at 35 m/s moves freely under the influence of gravity. How long does it take it to stop and begin falling downward? How high does it rise in this time?

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Solution

Solution: Velocity changes by 9.8 m/s every second; to change by the 35 m/s required come to a stop for an instant before falling back down (at which instant it is at its highest point), the time required is ( 35 m/s)/(9.8 m/s/s) = 3.571 seconds. The average velocity for this time will be ( 35 m/s + 0 m/s)/2 = 17.5 m/s. In 3.571 seconds, then, the object will fall ( 17.5 m/s)( 3.571 sec) = 62.49 meters.

Generalized Solution

Generalized Response: The time required for a projectile with initial upward vertical velocity v0y to come to rest is found by first observing that the change in velocity must be - v0y (using the upward direction as positive). In this case the acceleration of gravity will be -g, and the time required for the change in velocity will be `dt = - v0y / (-g) = v0y / g.

The height to which the object will rise will be the product of its average velocity and this time interval. The average velocity is vyAve = (v0y + 0) / 2 = v0y / 2, so the height to which the object will rise is

max height = `dsy = vyAve * `dt = (v0y / 2) * (v0y / g) = v0y^2 / (2g).

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Explanation in terms of Figure(s), Extension

Figure description:

Figure(s)