Topic 04 – Linear Kinematics (Part 2)
Lesson Learning Outcomes
- Define average acceleration.
- Define instantaneous acceleration.
- Name the units of measurement for acceleration.
- Use the equations of projectile motion to determine the vertical or horizontal position of a projectile given the initial velocities and time.
Guiding Questions
- Why is understanding the concept of acceleration in sports important?
- Why is determining the vertical or horizontal position of projectile important?
- How do you apply principles of projectile motions to sport activities?
Acceleration
Acceleration is the rate of change in velocity. Because velocity is a vector quantity, with a number and direction associated with it, acceleration is also a vector quantity, with a number and direction associated with it. An object accelerates if the magnitude or direction of its velocity changes.
Average acceleration is defined as the change in velocity divided by the time it took for that velocity change to take place. Mathematically, this is
Acceleration can be positive or negative. If the final velocity is less (slower) than the initial velocity, the change in velocity is a negative number, and the resulting average acceleration is negative. This happens if an object slows down in the positive direction. You may have thought of this as a deceleration, but we’ll call it a negative acceleration. A negative average acceleration will also result if the initial and final velocities are both negative and if the final velocity is a larger negative number than the initial velocity. This occurs if an object is speeding up in the negative direction. The units for describing acceleration are a unit of length divided by a unit of time divided by a unit of time. The SI units for describing acceleration are meters per second per second or meters per second squared (m/s2).
Instantaneous acceleration is the acceleration of an object at an instant in time. Instantaneous acceleration indicates the rate of change of velocity at that instant in time. Because acceleration is a vector (as are force, displacement, and velocity), it can also be resolved into component accelerations. This is true for both average and instantaneous accelerations.
Projectile Motion
A projectile is an object that has been projected into the air or dropped and is acted on only by the forces of gravity and air resistance. If air resistance is too small to measure, and the only force acting on a projectile is the force of the earth’s gravity, then the force of gravity will accelerate the projectile.
Consider throwing a ball straight up into the air where the direction along the y-axis is upward.
As the ball leaves your hand,
- velocity is positive (moving upward)
- acceleration is negative (ball is slowing down)
When the ball reaches the peak in the air,
- velocity change from positive to negative
- accleration is still negative (from upward to downward movement)
After the ball past its peak and falls downwards
- velocity is negative (moving downwards)
- acceleration is still negative (moving downward)
Despite the changing direction of the ball, its vertical acceleration was always downwards (negative) while it was in the air. The direction of its acceleration was constant. Since the ball’s weight does not change while it’s in the air, the net external force acting on the ball is constant and equal to the weight of the ball. This means that the acceleration of the ball is constant as well. In the previous chapter we learned that this acceleration, the acceleration due to gravity or g, is 9.81 m/s2 downward. This is a uniform acceleration.
Vertical Motion of Projectile
Since the vertical acceleration of the ball is constant, we already have one equation to describe this kinematic variable. If we define upward as the positive vertical direction, then a = g = –9.81 m/s2 (negative due to downward direction).
In our case we know the acceleration of the ball at any instant in time—it’s 9.81 m/s2 downward. But since the acceleration is constant, 9.81 m/s2 is also the average acceleration. So, we can substitute g for average acceleration
We can determine instantaneous vertical velocity of the ball (vf) at the end of sometime interval (𝚫t) if we know its initial vertical velocity (vi) and the length of the time interval.
If you remember your algebra, you might recognize this as the equation for a line:
y = mx + b
In the previous equation,
The vertical velocity of the ball is directly proportional to the time that the ball has been in the air. We can find out the vertical position of the ball by using the definition for average velocity (in the y-axis) from the football example earlier.
Since velocity is linearly proportional to time (it’s defined by a linear equation), the average velocity over a time interval is equal to the velocity midway between the initial and final velocities. This velocity is the average of the initial and final velocities. Combining the previous equations,
This equation gives us a means of determining the vertical position of the ball (yf) at the end of a time interval (𝚫t) if we know its initial vertical velocity (vi) and the length of the time interval. There is one more equation that describes vertical velocity of the ball as a function of its vertical displacement and initial vertical velocity.
We now have four equations to describe the vertical motion of a projectile.
Vertical position of projectile yf = yi + vi𝚫t + ½ g(𝚫t)2
Vertical velocity of projectile vf = vi + g𝚫t
v2 = v2 + 2g𝚫y
Vertical acceleration of projectile a = g = −9.81 m/s2
where
yi = initial vertical position,
yf = final vertical position,
𝚫y = yf − yi = vertical displacement,
𝚫t = change in time,
vi = initial vertical velocity,
vf = final vertical velocity, and
g = acceleration due to gravity = −9.81 m/s2.
If we are analyzing the motion of something that is dropped, the equations are simplified. For a dropped object, vf = 0. If we set the vertical scale to zero at the position the object was dropped from, then yi = 0 as well. For a dropped object, the equations become the following:
Vertical position of falling object yf = ½ g(𝚫t)2
Vertical velocity of falling object vf = g𝚫t
v2 = 2g𝚫y
Horizontal Motion of Projectile
Throw a ball in the air from one hand to the other so the ball has both vertical motion and horizontal motion. If we resolve the motion of the ball into horizontal (x) and vertical (y) components, we know that gravity is an external force that acts in the vertical direction and pulls downward on the ball. The only thing that could exert a horizontal force on the ball is the air through which the ball moves. This force will probably be very small in most cases, and its effect will be too small to notice. If air resistance is negligible, the horizontal velocity of the ball should not change from the time it leaves your hand until it contacts your other hand or another object, since no horizontal forces act on the ball. Its horizontal velocity is positive. If the ball doesn’t speed up or slow down or change direction, it is not accelerating in the horizontal direction.
Imagine watching a basketball game from above where all you see is the horizontal motion of the ball. (fig 2.7) To represent the motion, we show the position of the basketball at four instants in time, each 0.10 s apart. Notice that the images line up along a straight line, so the motion of the ball is in a straight line. Also notice that the displacement of the ball over each interval of time is the same, so the velocity of the ball is constant. The horizontal velocity of a projectile is constant, and its horizontal motion is in a straight line.
Now we can derive equations describing the horizontal position,velocity, and velocity, and acceleration of a projectile. We start with the fact that the horizontal velocity of a projectile is constant.
v = vf = vi = constant
If the horizontal velocity is constant, that means there is no change in horizontal velocity. If horizontal velocity doesn’t change, then horizontal acceleration must be zero, since acceleration was defined as the rate of change in velocity.
Also, if the horizontal velocity is constant, then the average horizontal velocity of the projectile is the same as its instantaneous horizontal velocity. Since average velocity is displacement divided by time, displacement is equal to velocity times time
If our measuring system is set up so that the initial horizontal position (xi) is zero, then equation simplifies to
x = v𝚫t
We now have the equations to describe the horizontal motion of a projectile.
Horizontal position of projectile xf = xi + v𝚫t
x = v𝚫t (if initial position is zero)
Horizontal velocity of projectile v = vf = vi = constant
Horizontal acceleration of projectile a = 0
where
xi = initial horizontal position,
xf = final horizontal position,
𝚫t = change in time,
vi = initial horizontal velocity, and
vf = final horizontal velocity.
Combined Horizontal and Vertical Motions of a Projectile
The vertical and horizontal motions of a projectile are independent of each other. In other words, a projectile continues to accelerate downward at 9.81 m/s2 with or without horizontal motion, and the horizontal velocity of a projectile remains constant even though the projectile is accelerating downward at 9.81 m/s2. Although the motions of a projectile are independent of each other, an equation can be derived to describe the path of a projectile in two dimensions.
Combining the two equations of verticle and horizontal position of projectile:
x = v𝚫t
yf = yi + vi𝚫t + ½ g(𝚫t)2
The equation above is the equation of a parabola. It describes the vertical (y) and horizontal (x) coordinates of a projectile during its flight based solely on the initial vertical position and vertical and horizontal velocities. The figure below shows the parabolic path followed by a ball thrown in the air with an initial vertical velocity of 6.95 m/s and an initial horizontal velocity of 4.87 m/s. The ball was photographed at a rate of 12 frames per second, so the position of the ball at each 0.0833 s interval is shown in the figure. Notice that the horizontal displacements over each time interval are the same and that the path is symmetrical on either side of the peak. The peak height actually occurs between the ninth and 10th ball images as counted from the left.
Projectile in Sports
In projectile activities, the initial conditions (the initial position and initial velocity) of the projectile determine the motion that the projectile will have. In sports involving projectiles, the athlete’s objective when throwing, kicking, striking, shooting, or hitting the projectile usually concerns one of three things: time of flight, peak height reached by the projectile, or horizontal displacement.
Time of flight of a projectile is dependent on two things: initial vertical velocity and initial vertical position. Maximizing time in the air is desirable in certain situations in sport such as a football punt or a lob in tennis. In these situations, the initial vertical velocity of the projectile is relatively large (compared to the horizontal velocity), and the angle of projection is above 45°. The optimal angle of projection to achieve maximum height and time of flight is 90° or straight up. In some sport activities, minimizing the projectile’s time in the air is important. Examples of these activities include a spike in volleyball, an overhead smash in tennis, throws in baseball, and a penalty kick in soccer. In these situations, the initial upward vertical velocity of the ball is minimized or the ball may even have an initial downward velocity. The projection angle is relatively small—less than 45°—and in some cases even less than zero.
The peak height reached by a projectile is also dependenton its initial height and initial vertical velocity. The higher a projectile is at release and the faster it is moving upward at release, the higher it will go. Maximizing peak height is important in sports such as volleyball and basketball, where the players themselves are the projectiles. Again, the athlete is the projectile. In these activities, the angle of projection is large, above 45°.
Maximizing the horizontal displacement or range of a projectile is the objective of several projectile sports. Examples of these include many of the field events in track and field, including the shot put, hammer throw, discus throw, javelin throw, and long jump. For the shot put, hammer throw, and long jump, air resistance is too small to significantly affect things, so our projectile equations are valid. Our analysis of these situations may require the use of equations. If we want to maximize horizontal displacement, then equation 2.24 may be useful.
𝚫x = v𝚫t
This equation describes horizontal displacement (𝚫x) as a function of initial horizontal velocity (v) and time (𝚫t). But time in this case would be total time in the air or the flight time of the projectile. We just saw that the flight time of a projectile is determined by its initial height and its initial vertical velocity. The horizontal displacement of a projectile is thus determined by three things: initial horizontal velocity, initial vertical velocity, and initial height. If the initial height of release is zero (the same as the landing height), then the resultant velocity (the sum of vertical and horizontal velocities) at release determines the horizontal displacement of the projectile. The faster you can throw something, the farther it will go.
If the initial velocity of the ball is totally vertical (a projection angle of 90°), the initial horizontal velocity (v in equation 2.24) would be zero, and the horizontal displacement would be zero as well. If the initial velocity of the ball is totally horizontal (a projection angle of zero), the flight time (𝚫t in equation 2.24) would be zero, and the horizontal displacement would be zero as well. Obviously, a combination of horizontal and vertical initial velocities (and a projection angle somewhere between 0° and 90°) would be better.
If the resultant velocity is the same no matter what the angle of projection, then maximum horizontal displacement will occur if the horizontal and vertical components of the initial velocity are equal, or when the projection angle is 45°. Horizontal displacement is determined by initial horizontal velocity and time in the air, but time in the air is determined by initial vertical velocity alone (if height of release is zero). It makes sense that these two variables—initial horizontal and vertical velocities— would have equal influence on horizontal displacement.
Summary
- To maximize the time of flight or the height reached by a projectile, the vertical component of release velocity should be maximized, and the projection angle should be above 45°.
- To minimize the time of flight of a projectile, the upward component of release velocity should be minimized, and the projection angle should be much lower than 45° and, in some situations, may even be below horizontal.
- To maximize the horizontal displacement of a projectile, release velocity should be maximized, and a higher release height is better.
- The horizontal component of release velocity should be slightly faster than the vertical component so that the projection angle is slightly lower than 45°.
- The higher the release height and the greater the lift effects of air resistance on the projectile, the farther below 45° the projection angle should be.