Determination of Horizontal Displacement
The horizontal displacement of a projectile is dependent upon the horizontal component of the initial velocity. As discussed in the previous part of this lesson, the horizontal displacement of a projectile can be determined using the equation
x = vx * t
If a projectile has a time of flight of 8 seconds and a horizontal velocity of 20 m/s, then the horizontal displacement is 160 meters (20 m/s*8 s). If a projectile has a time of flight of 8 seconds and a horizontal velocity of 34 m/s, then the projectile has a horizontal displacement of 272 meters (34 m/s*8 s). The horizontal displacement is dependent upon the only horizontal parameter which exists for projectiles - the horizontal velocity.
Determination of the Peak Height
A non-horizontally launched projectile with an initial vertical velocity of 40 m/s will reach its peak in 4 seconds. The process of rising to the peak is a vertical motion and is again dependent upon vertical motion parameters (the initial vertical velocity and the vertical acceleration). The height of the projectile at this peak position can be determined using the equation
y = viy * t + 0.5 * g * t2
where viy is the initial vertical velocity (in m/s), g is the acceleration of gravity (-10 m/s/s) and t is the time (in seconds) it takes to reach the peak. This equation can be successfully used to determine the vertical displacement of the projectile through the first half of its trajectory (i.e., peak height) provided that the algebra is properly performed and the proper values are substituted for the given variables. Special attention should be given to the facts that the t in the equation is the time up to the peak and the g has a negative value (-10 m/s/s).
Horizontally Launched Projectile Problems
One of the powers of physics is to be able to use physics principles to make predictions about the final outcome of a moving object. Such predictions are made through the application of physical principles and mathematical formulas to a given set of initial conditions. In the case of projectiles, a student of physics can use information about the initial velocity and position of a projectile to predict such things as how much time the projectile is in the air and how far the projectile will go. The physical principles which must be applied are those discussed previously in Lesson . The mathematical formula which are used are commonly referred to as kinematic equations. Combining the two allows one to make predictions concerning the motion of a projectiles. Such predictions are often made in response to a problem posed by a teacher, known as projectile problems.
There are two basic types of projectile problems which we will discuss in this course. While the general principles are the same for each type of problem, the approach will vary due to the fact the problems differ in terms of their initial conditions. The two types of problems are
A projectile is launched with a horizontal velocity from an elevated position and follows a parabolic path to the ground; predictable unknowns include the initial speed of the projectile, the initial height of the projectile, the time of flight, and the horizontal distance of the projectile.
Examples of this type of problem are
- A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location.
- A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.
A projectile is launched at an angle to the horizontal and rises upwards to a peak while moving horizontally; upon reaching the peak, the projectile falls with a motion which is symmetrical to its path upwards to the peak; predictable unknowns include the time of flight, the horizontal range, and the height of the projectile when it is at its peak.
Examples of this type of problem are
- A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the football.
- A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper.
The latter type of problem will be the subject of the next part of Lesson. In this part of Lesson, we will focus on the first type of problem - sometimes referred to as horizontally-launched projectile problems. Three common kinematic equations which will be employed for both type of problems include the following:
The above equations work well for motion in one-dimension, but a projectile is usually moving in two dimensions - both horizontally and vertically. Since these two components of motion are independent of each other, two distinctly separate sets of equations are needed - one for the projectile's horizontal motion and one for its vertical motion. Thus, the three equations above are transformed into two sets of three equations. For the horizontal components of motion, the equations are
Of these three equations, the top equation is the most commonly used. An application of projectile concepts to each of these equations would also lead one to conclude that any term with ax in it would cancel out of the equation since ax = 0 m/s/s.
For the vertical components of motion, the three equations are
In each of the above equations, the vertical acceleration of a projectile is known to be -10 m/s/s (the acceleration of gravity). Furthermore, for the special case of the first type of problem (horizontally-launched projectile problems), viy = 0 m/s; thus, any term with viy in it will cancel out of the equation.
The above sets of three equations are the kinematic equations which will be used to solve projectile motion problems.
To illustrate the usefulness of the above equations in making predictions about the motion of a projectile, consider their use in the solution of the following problem.
| Example
A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location. |
The solution of this problem begins with declaring the values of the known information in terms of the symbols of the kinematic equations - x, y, vix, viy, ax, ay, and t. In this case, the following information is either given or implicit in the problem statement:
| Horizontal Information | Vertical Information |
| x = ??? vix = 2.4 m/s ax = 0 m/s/s | y = -0.60 m viy = 0 m/s ay = -10 m/s/s |
As indicated in the table, the unknown quantity is the horizontal displacement (and the time of flight) of the pool ball. The solution of the problem now requires the selection of an appropriate strategy for using the kinematic equations and the known information to solve for the unknown quantities. It will almost always be the case that such a strategy demands that one of the vertical equations be used to determine the time of flight of the projectile and then one of the horizontal equations be used to find the other unknown quantities (or vice versa - first use the horizontal and then the vertical equation). A careful listing of known quantities (as in the table above) provides cues for the selection of the strategy. For example, the table above reveals that there is more vertical information known than horizontal information. Thus, it would be reasonable that a vertical equation be used with the vertical values to determine time and then the horizontal equations be used to determine "x." The first vertical equation (y = viy*t +0.5*ay*t2) will allow for the determination of the time. Once the appropriate equation has been selected, the physics problem becomes transformed into an algebra problem. By substitution of known values, the equation takes the form of
-0.60 m = (0 m/s)*t + 0.5*(-10 m/s/s)*t2
Since the first term on the right side of the equation reduces to 0, the equation can be simplified to
-0.60 m = (-5.0 m/s/s)*t2
If both sides of the equation are divided by -5.0 m/s/s, the equation becomes
0.12 s2 = t2
By taking the square root of both sides of the equation, the time of flight can then be determined.
t = 0.35 s (rounded from 0.3464 s)
Once the time has been determined, a horizontal equation can be used to determine the horizontal displacement of the pool ball. Recall from the given information, vix = 2.4 m/s and ax = 0 m/s/s. The first horizontal equation (x = vix*t + 0.5*ax*t2) can then be used to solve for "x." With the equation selected, the physics problem once more becomes transformed into an algebra problem. By substitution of known values, the equation takes the form of
x = (2.4 m/s)*(0.3464 s) + 0.5*(0 m/s/s)*(0.3464 s)2
Since the second term on the right side of the equation reduces to 0, the equation can then be simplified to
x = (2.4 m/s)*(0.3464 s)
Thus,
x = 0.83 m (rounded from 0.8313 m)
The answers to the stated problem is that the pool ball is in the air for 0.35 seconds and lands a horizontal distance of 0.83 m from the edge of the pool table.
The following procedure summarizes the above problem-solving approach.
- Carefully read the problem and list known and unknown information in terms of the symbols of the kinematic equations. For convenience sake, make a table with horizontal information on one side and vertical information on the other side.
- Identify the unknown quantity which the problem requests you to solve for.
- Select either a horizontal or vertical equation to solve for the time of flight of the projectile.
- With the time determined, use one of the other equations to solve for the unknown. (Usually, if a horizontal equation is used to solve for time, then a vertical equation can be used to solve for the final unknown quantity.)
One caution is in order: the sole reliance upon 4- and 5-step procedures to solve physics problems is always a dangerous approach. Physics problems are usually just that - problems! And problems can often be simplified by the use of short procedures as the one above. However, not all problems can be solved with the above procedure. While steps 1 and 2 above are critical to your success in solving horizontally-launched projectile problems, there will always be a problem which doesn't "fit the mold." Problem-solving is not like cooking; it is not a mere matter of following a recipe. Rather, problem-solving requires careful reading, a firm grasp of conceptual physics, critical thought and analysis, and lots of disciplined practice. Never divorce conceptual understanding and critical thinking from your approach to solving problems.
Check Your Understanding
A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.
Non-Horizontally Launched Projectile Problems
The use of kinematic equations to solve projectile problems was introduced and demonstrated. These equations were used to solve problems involving the launching of projectiles in a horizontal direction from an elevated position. In this section of Lesson, the use of kinematic equations to solve non-horizontally launched projectiles will be demonstrated. A non-horizontally launched projectile is a projectile which begins its motion with an initial velocity that is both horizontal and vertical. To treat such problems, the same principles which were discussed earlier in Lesson will have to be combined with the kinematic equations for projectile motion. You may recall from earlier, that there are two sets of kinematic equations - a set of equations for the horizontal components of motion and a similar set for the vertical components of motion. For the horizontal components of motion, the equations are
Of these three equations, the top equation is the most commonly used. The other two equations are seldom (if ever) used. An application of projectile concepts to each of these equations would also lead one to conclude that any term with ax in it would cancel out of the equation since ax = 0 m/s/s.
For the vertical components of motion, the three equations are
In each of the above equations, the vertical acceleration of a projectile is known to be -10 m/s/s (the acceleration of gravity).
As discussed earlier in Lesson, the vix and viy values in each of the above sets of kinematic equations can be determined by the use of trigonometric functions. The initial x-velocity (vix) can be found using the equation vix = vi*cosine(Theta) where Theta is the angle which the velocity vector makes with the horizontal. The initial y-velocity (viy) can be found using the equation viy = vi*sine(Theta) where Theta is the angle which the velocity vector makes with the horizontal. The topic of components of the velocity vector was discussed earlier in Lesson 2.
To illustrate the usefulness of the above equations in making predictions about the motion of a projectile, consider their use in the solution of the following problem.
| Example A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal displacement, and the peak height of the football. |
The solution of any non-horizontally launched projectile problem (in which vi and Theta are given) should begin by first breaking the initial velocity into horizontal and vertical components using the trigonometric functions discussed above. Thus,
| Horizontal Component | Vertical Component |
| vix = vi*cos(Theta) vix = 25 m/s*cos(45 deg) vix = 17.7 m/s | viy = vi*sin(Theta) viy = 25 m/s*sin(45 deg) viy = 17.7 m/s |
In this case, it happens that the vix and the viy values are the same (as will always be the case when the angle is 45-degrees).
The solution continues by declaring the values of the known information in terms of the symbols of the kinematic equations - x, y, vix, viy, ax, ay, and t. In this case, the following information is either given or implicit in the problem statement: | Horizontal Information | Vertical Information |
| x = ??? vix = 17.7 m/s vfx = 17.7 m/s ax = 0 m/s/s | y = ??? viy = 17.7 m/s vfy = -17.7 m/s ay = -10 m/s/s |
As indicated in the table, the final x-velocity (vfx) is the same as the initial x-velocity (vix). This is due to the fact that the horizontal velocity of a projectile is constant; there is no horizontal acceleration. The table also indicates that the final y-velocity (vfy) has the same magnitude and the opposite direction as the initial y-velocity (viy). This is due to the symmetrical nature of a projectile's trajectory.
The unknown quantities are the horizontal displacement, the time of flight, and the height of the football at its peak. The solution of the problem now requires the selection of an appropriate strategy for using the kinematic equations and the known information to solve for the unknown quantities. There are a variety of possible strategies for solving the problem. A careful listing of known quantities (as in the table above) provides cues for the selection of a useful strategy.
From the vertical information in the table above and the second equation listed among the vertical kinematic equations (vfy = viy + ay*t), it becomes obvious that the time of flight of the projectile can be determined. By substitution of known values, the equation takes the form of
-17.7 m/s = 17.7 m/s + (-10 m/s/s)*t
The physics problem now takes the form of an algebra problem. By subtracting 17.7 m/s from each side of the equation, the equation becomes
-35.4 m/s = (-10 m/s/s)*t
If both sides of the equation are divided by -10 m/s/s, the equation becomes
3.54 s = t
The total time of flight of the football is 3.54 seconds.
With the time determined, information in the table and the horizontal kinematic equations can be used to determine the horizontal displacement (x) of the projectile. The first equation (x = vix*t + 0.5*ax*t2) listed among the horizontal kinematic equations is suitable for determining "x." With the equation selected, the physics problem once more becomes transformed into an algebra problem. By substitution of known values, the equation takes the form of
x = (17.7 m/s)*(3.54 s) + 0.5*(0 m/s/s)*(3.54 s)2
Since the second term on the right side of the equation reduces to 0, the equation can then be simplified to
x = (17.7 m/s)*(3.54 s) Thus,
x = 62.6 m
The horizontal displacement of the projectile is 62.6 m.
Finally, the problem statement asks for the height of the projectile at is peak. This is the same as asking "what is the vertical displacement (y) of the projectile when it is halfway through its trajectory?" In other words, find y when t = 1.77 seconds (one-half of the total time). To determine the peak height of the projectile (y with t = 1.77 sec), the first equation (y = viy*t +0.5*ay*t2) listed among the vertical kinematic equations can be used. By substitution of known values into this equation, it takes the form of
y = (17.7 m/s)*(1.77 s) + 0.5*(-10 m/s/s)*(1.77 s)2
Using a calculator, this equation can be simplified to
y = 31.3 m + (-15.7 m) And thus,
y = 15.6 m
The solution to the problem statement yields the following answers: the time of flight of the football is 3.54 s, the horizontal displacement of the football is 62.6 m, and the peak height of the football 15.6 m.
The following procedure summarizes the above problem-solving approach.
- Use the given values of the initial velocity (the magnitude and the angle) to determine the horizontal and vertical components of the velocity (vix and viy).
- Carefully read the problem and list known and unknown information in terms of the symbols of the kinematic equations. For convenience sake, make a table with horizontal information on one side and vertical information on the other side.
- Identify the unknown quantity which the problem requests you to solve for.
- Select either a horizontal or vertical equation to solve for the time of flight of the projectile. For non-horizontally-launched projectiles, the second equation listed among the vertical equations (vfy = viy + ay*t) is usually the most useful equation.
- With the time determined, use a horizontal equation (usually x = vix*t + 0.5*ax*t2 ) to determine the horizontal displacement of the projectile.
- Finally, the peak height of the projectile can be found using a time value which one-half the total time of flight. The most useful equation for this is usually y = viy*t +0.5*ay*t2 .
One caution is in order: the sole reliance upon 4- and 5-step procedures to solve physics problems is always a dangerous approach. Physics problems are usually just that - problems! And problems can often be simplified by the use of short procedures as the one above. However, not all problems can be solved with the above procedure. While steps 1, 2 and 3 above are critical to your success in solving non-horizontally launched projectile problems, there will always be a problem which doesn't "fit the mold." Problem-solving is not like cooking; it is not a mere matter of following a recipe. Rather, problem-solving requires careful reading, a firm grasp of conceptual physics, critical thought and analysis, and lots of disciplined practice. Never divorce conceptual understanding and critical thinking from your approach to solving problems.
Check Your Understanding
A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper.
Moderation Team
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