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.
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 onedimensional 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} } } {}.
y
1
=
0
+
13
.
0 m/s
1
.
00 s
+
1
2
−
9
.
80
m/s
2
1
.
00 s
2
=
8
.
10
m
y
1
=
0
+
13
.
0 m/s
1
.
00 s
+
1
2
−
9
.
80
m/s
2
1
.
00 s
2
=
8
.
10
m
size 12{y"" lSub { size 8{1} } =0+ left ("13" "." "0 m/s" right ) left (1 "." "00 s" right )+ { {1} over {2} } left (  9 "." "80"" m/s" rSup { size 8{2} } right ) left (1 "." "00 s" right ) rSup { size 8{2} } =8 "." "10"`m} {}
(5)
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.
v
1
=
v
0
−
gt
=
13
.
0 m/s
−
9
.
80 m/s
2
1
.
00 s
=
3
.
20 m/s
v
1
=
v
0
−
gt
=
13
.
0 m/s
−
9
.
80 m/s
2
1
.
00 s
=
3
.
20 m/s
size 12{v rSub { size 8{1} } =v rSub { size 8{0} }  ital "gt"="13" "." "0 m/s"  left (9 "." "80 m/s" rSup { size 8{2} } right ) left (1 "." "00 s" right )=3 "." "20 m/s"} {}
(6)
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.
Table 1: Results
Time, t

Position, y

Velocity, v

Acceleration, a

1
.
00 s
1
.
00 s
size 12{1 "." "00 s"} {}

8
.
10 m
8
.
10 m
size 12{8 "." "10 m"} {}

3
.
20 m/s
3
.
20 m/s
size 12{3 "." "20 m/s"} {}

−
9
.
80 m/s
2
−
9
.
80 m/s
2
size 12{9 "." "80 m/s" rSup { size 8{2} } } {}

2
.
00 s
2
.
00 s
size 12{2 "." "00 s"} {}

6
.
40 m
6
.
40 m
size 12{6 "." "40 m"} {}

−
6
.
60 m/s
−
6
.
60 m/s
size 12{  6 "." "60 m/s"} {}

−
9
.
80 m/s
2
−
9
.
80 m/s
2
size 12{9 "." "80 m/s" rSup { size 8{2} } } {}

3
.
00 s
3
.
00 s
size 12{3 "." "00 s"} {}

−
5
.
10 m
−
5
.
10 m
size 12{  5 "." "10 m"} {}

−
16
.
4 m/s
−
16
.
4 m/s
size 12{  "16" "." "4 m/s"} {}

−
9
.
80 m/s
2
−
9
.
80 m/s
2
size 12{9 "." "80 m/s" rSup { size 8{2} } } {}

Graphing the data helps us understand it more clearly.
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 freefall 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 freefall while arcing up as well as down, as we will discuss in more detail later.
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