A person standing on the edge of a high cliff throws a rock straight up with an initial velocity of 13.0 m/s. The rock misses the edge of the cliff as it falls back to earth. Calculate the position and velocity of the rock 1.00 s, 2.00 s, and 3.00 s after it is thrown, neglecting the effects of air resistance.

Strategy

Draw a sketch.

Figure 2

We are asked to determine the position yy size 12{y} {} at various times. It is reasonable to take the initial position y0y0 size 12{y rSub { size 8{0} } } {} to be zero. This problem involves one-dimensional motion in the vertical direction. We use plus and minus signs to indicate direction, with up being positive and down negative. Since up is positive, and the rock is thrown upward, the initial velocity must be positive too. The acceleration due to gravity is downward, so aa size 12{a} {} is negative. It is crucial that the initial velocity and the acceleration due to gravity have opposite signs. Opposite signs indicate that the acceleration due to gravity opposes the initial motion and will slow and eventually reverse it.

Since we are asked for values of position and velocity at three times, we will refer to these as y1y1 size 12{y rSub { size 8{1} } } {} and v1v1 size 12{v rSub { size 8{1} } } {}; y2y2 size 12{y rSub { size 8{2} } } {} and v2v2 size 12{v rSub { size 8{2} } } {}; and y3y3 size 12{y rSub { size 8{3} } } {} and v3v3 size 12{v rSub { size 8{3} } } {}.

Solution for Position y 1 y 1 size 12{y rSub { size 8{1} } } {}

1. Identify the knowns. We know that y0=0y0=0 size 12{y rSub { size 8{0} } =0} {}; v0=13.0 m/sv0=13.0 m/s size 12{v rSub { size 8{0} } ="13" "." "0 m/s"} {}; a=−g=−9.80 m/s2a=−g=−9.80 m/s2 size 12{a= - g= - 9 "." "80 m/s" rSup { size 8{2} } } {}; and t=1.00 st=1.00 s size 12{t=1 "." "00 s"} {}.

2. Identify the best equation to use. We will use y=y0+v0t+12at2y=y0+v0t+12at2 size 12{y=y rSub { size 8{0} } +v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } } {} because it includes only one unknown, yy size 12{y} {} (or y1y1 size 12{y rSub { size 8{1} } } {}, here), which is the value we want to find.

3. Plug in the known values and solve for y1y1 size 12{y rSub { size 8{1} } } {}.

Discussion

The rock is 8.10 m above its starting point at t=1.00t=1.00 size 12{t=1 "." "00"} {} s, since y1>y0y1>y0 size 12{y rSub { size 8{1} } >y rSub { size 8{0} } } {}. It could be moving up or down; the only way to tell is to calculate v1v1 size 12{v rSub { size 8{1} } } {} and find out if it is positive or negative.

Solution for Velocity v 1 v 1 size 12{v rSub { size 8{1} } } {}

1. Identify the knowns. We know that y0=0y0=0 size 12{y rSub { size 8{0} } =0} {}; v0=13.0 m/sv0=13.0 m/s size 12{v rSub { size 8{0} } ="13" "." "0 m/s"} {}; a=−g=−9.80 m/s2a=−g=−9.80 m/s2 size 12{a= - g= - 9 "." "80 m/s" rSup { size 8{2} } } {}; and t=1.00 st=1.00 s size 12{t=1 "." "00 s"} {}. We also know from the solution above that y1=8.10 my1=8.10 m size 12{y rSub { size 8{1} } =8 "." "10 m"} {}.

2. Identify the best equation to use. The most straightforward is v=v0−gtv=v0−gt size 12{v=v rSub { size 8{0} } - ital "gt"} {} (from v=v0+atv=v0+at size 12{v=v rSub { size 8{0} } + ital "at"} {}, where a=gravitational acceleration=−ga=gravitational acceleration=−g size 12{a="gravitational acceleration"= - g} {}).

3. Plug in the knowns and solve.

Discussion

The positive value for v1v1 means that the rock is still heading upward at t=1.00st=1.00s. However, it has slowed from its original 13.0 m/s, as expected.

Solution for Remaining Times

The procedures for calculating the position and velocity at t=2.00st=2.00s size 12{t=2 "." "00"`s} {} and 3.00 s3.00 s size 12{3 "." "00 s"} {} are the same as those above. The results are summarized in Table 1 and illustrated in Figure 3.

Graphing the data helps us understand it more clearly.

Figure 3: Vertical position, vertical velocity, and vertical acceleration vs. time for a rock thrown vertically up at the edge of a cliff. Notice that velocity changes linearly with time and that acceleration is constant. Misconception Alert! Notice that the position vs. time graph shows vertical position only. It is easy to get the impression that the graph shows some horizontal motion—the shape of the graph looks like the path of a projectile. But this is not the case; the horizontal axis is time, not space. The actual path of the rock in space is straight up, and straight down.

Discussion

The interpretation of these results is important. At 1.00 s the rock is above its starting point and heading upward, since y1y1 size 12{y rSub { size 8{1} } } {} and v1v1 size 12{v rSub { size 8{1} } } {} are both positive. At 2.00 s, the rock is still above its starting point, but the negative velocity means it is moving downward. At 3.00 s, both y3y3 size 12{y rSub { size 8{3} } } {} and v3v3 size 12{v rSub { size 8{3} } } {} are negative, meaning the rock is below its starting point and continuing to move downward. Notice that when the rock is at its highest point (at 1.5 s), its velocity is zero, but its acceleration is still −9.80 m/s2−9.80 m/s2 size 12{-9 "." "80 m/s" rSup { size 8{2} } } {}. Its acceleration is −9.80 m/s2−9.80 m/s2 size 12{-9 "." "80 m/s" rSup { size 8{2} } } {} for the whole trip—while it is moving up and while it is moving down. Note that the values for yy size 12{y} {} are the positions (or displacements) of the rock, not the total distances traveled. Finally, note that free-fall applies to upward motion as well as downward. Both have the same acceleration—the acceleration due to gravity, which remains constant the entire time. Astronauts training in the famous Vomit Comet, for example, experience free-fall while arcing up as well as down, as we will discuss in more detail later.

What happens if the person on the cliff throws the rock straight down, instead of straight up? To explore this question, calculate the velocity of the rock when it is 5.10 m below the starting point, and has been thrown downward with an initial speed of 13.0 m/s.

Strategy

Draw a sketch.

Figure 4

Since up is positive, the final position of the rock will be negative because it finishes below the starting point at y0=0y0=0 size 12{y rSub { size 8{0} } =0} {}. Similarly, the initial velocity is downward and therefore negative, as is the acceleration due to gravity. We expect the final velocity to be negative since the rock will continue to move downward.

Solution

1. Identify the knowns. y0=0y0=0; y1=−5.10 my1=−5.10 m; v0=−13.0 m/sv0=−13.0 m/s; a=−g=−9.80 m/s2a=−g=−9.80 m/s2 size 12{a= - g= - 9 "." "80"" m/s" rSup { size 8{2} } } {}.

2. Choose the kinematic equation that makes it easiest to solve the problem. The equation v2=v02+2a(y−y0)v2=v02+2a(y−y0) works well because the only unknown in it is vv. (We will plug y1y1 in for yy.)

3. Enter the known values

where we have retained extra significant figures because this is an intermediate result.

Taking the square root, and noting that a square root can be positive or negative, gives

The negative root is chosen to indicate that the rock is still heading down. Thus,

Discussion

Note that this is exactly the same velocity the rock had at this position when it was thrown straight upward with the same initial speed. (See Example 1 and Figure 5(a).) This is not a coincidental result. Because we only consider the acceleration due to gravity in this problem, the speed of a falling object depends only on its initial speed and its vertical position relative to the starting point. For example, if the velocity of the rock is calculated at a height of 8.10 m above the starting point (using the method from Example 1) when the initial velocity is 13.0 m/s straight up, a result of ±3.20 m/s±3.20 m/s size 12{ +- 3 "." "20"`"m/s"} {} is obtained. Here both signs are meaningful; the positive value occurs when the rock is at 8.10 m and heading up, and the negative value occurs when the rock is at 8.10 m and heading back down. It has the same speed but the opposite direction.

Figure 5: (a) A person throws a rock straight up, as explored in Example 1. The arrows are velocity vectors at 0, 1.00, 2.00, and 3.00 s. (b) A person throws a rock straight down from a cliff with the same initial speed as before, as in Example 2. Note that at the same distance below the point of release, the rock has the same velocity in both cases.

Another way to look at it is this: In Example 1, the rock is thrown up with an initial velocity of 13.0 m/s13.0 m/s. It rises and then falls back down. When its position is y=0y=0 on its way back down, its velocity is −13.0 m/s−13.0 m/s. That is, it has the same speed on its way down as on its way up. We would then expect its velocity at a position of y=−5.10 my=−5.10 m to be the same whether we have thrown it upwards at +13.0 m/s+13.0 m/s or thrown it downwards at −13.0 m/s−13.0 m/s. The velocity of the rock on its way down from y=0y=0 is the same whether we have thrown it up or down to start with, as long as the speed with which it was initially thrown is the same.

The acceleration due to gravity on Earth differs slightly from place to place, depending on topography (e.g., whether you are on a hill or in a valley) and subsurface geology (whether there is dense rock like iron ore as opposed to light rock like salt beneath you.) The precise acceleration due to gravity can be calculated from data taken in an introductory physics laboratory course. An object, usually a metal ball for which air resistance is negligible, is dropped and the time it takes to fall a known distance is measured. See, for example, Figure 6. Very precise results can be produced with this method if sufficient care is taken in measuring the distance fallen and the elapsed time.

Figure 6: Positions and velocities of a metal ball released from rest when air resistance is negligible. Velocity is seen to increase linearly with time while displacement increases with time squared. Acceleration is a constant and is equal to gravitational acceleration.

Suppose the ball falls 1.0000 m in 0.45173 s. Assuming the ball is not affected by air resistance, what is the precise acceleration due to gravity at this location?

Strategy

Draw a sketch.

Figure 7

We need to solve for acceleration aa size 12{a} {}. Note that in this case, displacement is downward and therefore negative, as is acceleration.

Solution

1. Identify the knowns. y0=0y0=0; y=–1.0000 my=–1.0000 m; t=0.45173t=0.45173; v0=0v0=0 size 12{v rSub { size 8{0} } =0} {}.

2. Choose the equation that allows you to solve for aa size 12{a} {} using the known values.

3. Substitute 0 for v0v0 size 12{v rSub { size 8{0} } } {} and rearrange the equation to solve for aa size 12{a} {}. Substituting 0 for v0v0 size 12{v rSub { size 8{0} } } {} yields

Solving for aa size 12{a} {} gives

4. Substitute known values yields

so, because a=−ga=−g size 12{a= - g} {} with the directions we have chosen,

Discussion

The negative value for aa size 12{a} {} indicates that the gravitational acceleration is downward, as expected. We expect the value to be somewhere around the average value of 9.80 m/s29.80 m/s2 size 12{9 "." "80 m/s" rSup { size 8{2} } } {}, so 9.8010 m/s29.8010 m/s2 size 12{9 "." "8010 m/s" rSup { size 8{2} } } {} makes sense. Since the data going into the calculation are relatively precise, this value for gg size 12{g} {} is more precise than the average value of 9.80 m/s29.80 m/s2 size 12{9 "." "80 m/s" rSup { size 8{2} } } {}; it represents the local value for the acceleration due to gravity.