Question #3, p. 35
On Mars, the acceleration due to gravity is -3.71 m/s squared. What would a
rock’s velocity be 3 s after you dropped it on Mars?
Since Mars has very thin atmosphere,
there is no significant air resistance. The
only force acting on the object is the
x0 = 0m
Welcome
gravitational force of the planet Mars.
v0 = 0 m/s
to Mars
Therefore the net force exerted on the
rock
rock is equal to the gravitational force. In
gy = -3.71 m/s2
three seconds the rock will move far
(toward the center of the planet),
therefore we can assume that the net
(gravitational) force acting on the rock is constant throughout the entire motion. Hence
from Newton’s second law, at the surface of Mars, the free fall acceleration vector g is
directed toward the center of Mars and its magnitude has a value of g = 3.71 m/s2.
x
Solving this problem is pretty much just like the
examples we have done so far for falling objects on
Equations
Earth. The reference frame of Mars, with the vertical
Acceleration
x-axis directed upward, can be used for the analysis a = -g
of the motion of the rock. The rock moves with a
constant acceleration. At the very moment the rock is
Velocity
released (t0 = 0s), its (initial) velocity will start at v = v + (a · t)
0
zero and consistently with the above figure, will be
increasing in the downward direction at rate of 3.71
Position
m/s per second. The vertical components of x = x + (v · t) + (½ · a · t2)
0
0
acceleration, velocity and position obey the included
equations. From the equation for velocity of an object in motion with constant
acceleration, the (vertical component of) velocity of the rock, at time 3s after being
released is
Present velocity (at 3s) = initial + acceleration * time = 0 - 3.71 m/s2 * 3 s = -11.13 m/s
Three seconds after a rock is dropped on Mars its velocity will be 11.13 m/s
downward (with both horizontal components of velocity being 0 m/s).
Question #4, p. 35
How far would a rock fall in 3 seconds if you dropped it on Mars?
As with the previous question, we are assuming that air resistance will not be a
significant factor in the calculation. This leaves only the gravitational interaction between
Mars and the rock. The (magnitude of the) acceleration due to gravity (a) on Mars is
3.71m/s2 will also apply to the falling rock. The time (t) instant being investigated is 3 s
after the rock has been dropped.
The same reference frame of Mars can be used for the analysis of the motion of the
falling rock. However, this time we will consider the position (x). Consistently with the
choice of the reference frame, (the vertical components of) the rock's initial position (x0)
and velocity (v0) are zero, While the (vertical component of the) acceleration is 3.71m/s2. Thus from the equation for position in a motion with constant acceleration, the
(vertical component of) position of the rock, at time 3s after being released is
P r e s e n t po s i t i on = i n i t i a l p o s i t i on + i n i t i a l v e l o c i t y * t i m e
+ ( 1 / 2 ) * a c c e l e r at i o n * ( t i m e ) 2 =
= 0m + 0m / s · 0 s + ( 1/ 2 ) * ( - 3 . 7 1m / s 2 ) * 3 s  - 1 6 . 7 m
Three seconds after a rock is dropped on Mars its position will 16.7m below the
initial location (both horizontal components of the position of the rock have 0m value).
By choosing a reference frame with the initial condition x0 = 0m, one can easily find the
displacement of the rock in three second. From the initial location, the rock would fall
16.7 m in three second. The analysis of the entire motion is presented in the following
table
Time
Velocity
Acceleration
Position
0s
1s
0 m/s
-3.7 m/s
-3.71 m/s2
-3.71 m/s2
0m
- 1.9 m
2s
-7.4 m/s
-3.71 m/s2
- 7.4 m
3s
-11.1 m/s
-3.71 m/s2
-16.7 m
Question #6 & #7, p. 35
A basketball player can leap upward .5 m. What is his initial velocity at the
start of the leap?
How long does the basketball player in Problem 6 remain in the air?
time
x
1.0 m
acceleration
velocity
position
0
-9.8
0
0.5m
?
-9.8
?
0
0.5m
0.5m
Ground
When the basketball player leaps off the ground, he leaves the ground with a large
upward velocity (large vertical component of the velocity). As soon as the basketball
player leaves the ground, only the gravitational force is exerted by the earth on the player.
Therefore, he has a downward acceleration of 9.8 m/s2. Consistently with the above
figure, the vertical component of the acceleration will have value of -9.8 m/s2. His initial
upward velocity (vertical component of the velocity) will gradually decrease until it
reaches 0. He then will begin to fall and his velocity will begin to increase in that
direction (the vertical component will get more and more negative) until he returns to the
ground.
In a motion with constant acceleration, the position of the object is a quadratic
function of time, determined by the initial position, initial velocity and the acceleration.
The solution is simplest if we choose the instant when the player is at the highest point as
the reference for time t0 = 0s. Consistently with all assumptions, the above table lists the
information give in the problem.
“present” height = “initial” height + “initial” velocity + ½ * acceleration * time2
Hence
0 m = 0.5 m + 0 m/s * t + ½ * (-9.8 m/s2) * t2
Solving for the unknown time (instant) t
4.9 m/s2 * t2 = 0.5 m
t2 = 0.5 m / (4.9 m/s2)  0.1 s2
Take the square root of both sides and you get two solutions for time when the player is
on the ground
t1 = - 0.32 s
and
t2 = 0.32 s
The negative value of time corresponds to the instant when the player started the leap (he
left the ground 0.32 s before he reached the highest point) and the positive identifies the
instant when the player finished the leap (he touched the ground 0.32 s after he reached
the highest point).
In a motion with constant acceleration, the velocity of the object is a linear function
of time, determined by the initial velocity and the acceleration
“present” velocity = “initial” velocity + acceleration * time
Hence at the time the player left the ground his upward velocity (vertical component) was
v = 0 m/s + (-9.8 m/s2) * (-0.32s) = 3.14m/s
At the start of the leap, the initial upward velocity (vertical component of velocity) of the
player is 3.14 m/s.
We have found that the player begins his leap 0.32 s before he reaches the highest
point and ends the leap 0.32 s before he reaches the highest point. (The way up is the
same amount of time as the way down.) Hence the total time he stays in air is
t = 0.32 s + 0.32 s = 0.64 s
One could also find that amount of time (length of the time interval) by subtracting the
time (instant) of the beginning of the leap (-0.32s) from the time (instant) of the end of
the leap (0.32 s)
t = 0.32 s – (-0.32 s) = 0.64 s
The basketball player stays in air for 0.64 s. If he jumps 0.5 m up, he does not even stay
in the air for a whole second.