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 total time of flight of the football is 3.61 seconds. If both sides of the equation are divided by -9.8 m/s/s, the equation becomes 3.61 s = t By subtracting 17.7 m/s from each side of the equation, the equation becomes -35.4 m/s = (-9.8 m/s/s) The physics problem now takes the form of an algebra problem. By substitution of known values, the equation takes the form of -17.7 m/s = 17.7 m/s + (-9.8 m/s/s) An organized listing of known quantities in two columns of a table provides clues for the selection of a useful strategy.įrom the vertical information in the table above and the second equation listed among the vertical kinematic equations (v fy = v iy + a y*t), it becomes obvious that the time of flight of the projectile can be determined. There are a variety of possible strategies for solving the problem. 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. The unknown quantities are the horizontal displacement, the time of flight, and the height of the football at its peak. This is due to the symmetrical nature of a projectile's trajectory. The table also indicates that the final y-velocity (v fy) has the same magnitude and the opposite direction as the initial y-velocity (v iy). This is due to the fact that the horizontal velocity of a projectile is constant there is no horizontal acceleration. In this case, the following information is either explicitly given or implied in the problem statement:Īs indicated in the table, the final x-velocity (v fx) is the same as the initial x-velocity (v ix). The solution continues by declaring the values of the known information in terms of the symbols of the kinematic equations - x, y, v ix, v iy, a x, a y, and t. In this case, it happens that the v ix and the v iy values are the same as will always be the case when the angle is 45-degrees. The solution of any non-horizontally launched projectile problem (in which v i and Theta are given) should begin by first resolving the initial velocity into horizontal and vertical components using the trigonometric functions discussed above. Determine the time of flight, the horizontal displacement, and the peak height of the football. 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.Ī football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. The topic of components of the velocity vector was discussed earlier in Lesson 2. sine(Theta) where Theta is the angle that the velocity vector makes with the horizontal.The initial y-velocity (v iy) can be found using the equation v iy = v i cosine(Theta) where Theta is the angle that the velocity vector makes with the horizontal.The initial x-velocity (v ix) can be found using the equation v ix = v i In each of the above equations, the vertical acceleration of a projectile is known to be -9.8 m/s/s (the acceleration of gravity).Īs discussed earlier in Lesson 2, the v ix and v iy values in each of the above sets of kinematic equations can be determined by the use of trigonometric functions. An application of projectile concepts to each of these equations would also lead one to conclude that any term with a x in it would cancel out of the equation since a x = 0 m/s/s.įor the vertical components of motion, the three equations are The other two equations are seldom (if ever) used. Of these three equations, the top equation is the most commonly used. For the horizontal components of motion, the equations are 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. To treat such problems, the same principles that were discussed earlier in Lesson 2 will have to be combined with the kinematic equations for projectile motion. A non-horizontally launched projectile is a projectile that begins its motion with an initial velocity that is both horizontal and vertical. In this section of Lesson 2, the use of kinematic equations to solve non-horizontally launched projectiles will be demonstrated. These equations were used to solve problems involving the launching of projectiles in a horizontal direction from an elevated position. In the previous part of Lesson 2, the use of kinematic equations to solve projectile problems was introduced and demonstrated.
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