One Dimensional Trajectories


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This lesson deals with calculation of position and velocity for a moving object in one dimension (we will consider two dimensions in the next lesson). For simplicity, we will consider an object which is constrained to move only up and down, such as the famous carnival show Test-Your-Strength attraction (often called a "high-striker") where one slams a sledgehammer on a board to drive a weight (called the traveller) up a pole in (a normally futile) attempt to ring a bell.

In this case, there are two forces acting on the weight; the initial velocity imparted by the sledge hammer (which is transmitted to the traveller via the lever of the board), and the force of gravity which pulls down on the traveller with a constant acceleration. In this case, we will ignore the friction of the traveller as it slides up the pole. We call the initial velocity V0.

Recall from previous lessons that a body in motion tends to stay in motion. Thus we can think of the traveller as always having the same velocity imparted by the sledge, it is just that the acceleration of gravity acts in the opposite direction and becomes larger, so the weight falls back down. The velocity at any given moment will be equal to that in intial velocity less the velocity due to the acceleration of gravity. From the previous lesson, recall that the velocity of an object is the product of the acceleration times the length of the acceleration.

So if we call time, t, and repreent the force of gravity as g, we can find the velocity due to gravity Vg ) at any time t as

Vg = g * t

so the velocity at any time t is

Vt = V0 - g * t

So, given a intial velocity, how high will the traveller travel? We solve this by observing that at the maximum height, the net velocity will be zero, so first we figure out long it takes to get to that point:

0 = V0 - g * t

thus

g * t = V0

and, finally,

t = V0 / g

Once we know at what time, t, we can easily figure out where it is. Again, from the previous lesson, recall that the distance an object falls can be calculated as

d = 1/2 * a * t2

Since the traveller has to fall back in the same time as it rose, we can then find the answer by subsituting gravity for the acceleration and our value of t from our earlier calculation.

Questions:

  1. If the initial velocity is 5 m/s, how long will it take to reach the zenith and what will be the height? For 8 m/s?
  2. If the traveller reaches a height of 8 meters, what was the initial velocity? How long did it take to get there?