Analysis of projectile motion There is a very useful aspect of two dimensional motion that can be used with great effect. Two dimensional motion can be resolved in to two linear motions in two mutually perpendicular directions : (i) one along horizontal direction and (ii) vertical direction. The linear motion in each direction can, then, be analyzed with the help of equivalent scalar system, in which sign of the attributes of the motion represent direction. We can analyze the projectile motion with the help of equations of motion. As the motion occurs in two dimensions, we need to use vector equations and interpret them either graphically or algebraically as per the vector rules. We know that algebraic methods consisting of component vectors render vectors analysis in relatively simpler way. Still, vector algebra requires certain level of skills to manipulate vector components in two directions. In the nutshell, the study of projectile motion is equivalent to two independent linear motions. This paradigm simplifies the analysis of projectile motion a great deal. Moreover, this equivalent construct is not solely a mathematical construct, but is physically verifiable fact. The motions in vertical and horizontal directions are indeed independent of each other. To illustrate this, let us consider the flight of two identical balls, which are initiated in motion at the same time. One ball is dropped vertically and another is projected in horizontal direction with some finite velocity from the same height. It is found that both balls reach the ground at the same time and also their elevations above the ground are same at all times during the motion.

Projectile motion Comparing vertical and projectile motion The fact that two balls reach the ground simultaneously and that their elevations from the ground during the motion at all times are same, point to the important aspect of the motion that vertical motion in either of the two motions are identical. This implies that the horizontal motion of the second ball does not interfere with its vertical motion. By extension, we can also say that the vertical motion of the second ball does not interfere with its horizontal motion.

Projectile motion and equations of motion Here, we describe the projectile motion with the help of a two dimensional rectangular coordinate system such that (This not not a requirement. One can choose reference coordinate system to one's convenience): Origin of the coordinate system coincides with point of projection. The x – axis represents horizontal direction. The y – axis represents vertical direction.

Initial velocity Let us consider that the projectile is thrown with a velocity “u” at an angle θ from the horizontal direction as shown in the figure. The component of initial velocity in the two directions are : u x = u cos θ u y = u sin θ

Projectile motion Component projection velocities in x and y directions

Equations of motion in vertical direction Motion in vertical direction is moderated by the constant force due to gravity. This motion, therefore, is described by one dimensional equations of motion for constant acceleration.

Velocity The velocity in the vertical direction is given by : v y = u y - g t An inspection of equation - 2 reveals that this equation can be used to determine velocity in vertical direction at a given time “t” or to determine time of flight “t”, if final vertical velocity is given. This assumes importance as we shall see that final vertical velocity at the maximum height becomes zero.

Projectile motion Vertical component of velocity during motion The equation for velocity further reveals that the magnitude of velocity is reduced by an amount “gt” after a time interval of “t” during upward motion. The projectile is decelerated in this part of motion (velocity and acceleration are in opposite direction). The reduction in the magnitude of velocity with time means that it becomes zero corresponding to a particular value of “t”. The vertical elevation corresponding to the position, when projectile stops, is maximum height that projectile attains. For this situation ( v y = 0), the time of flight “t” is obtained as : v y = u y - g t ⇒ 0 = u y - g t ⇒ t = u y g Immediately thereafter, projectile is accelerated in vertically downward direction with increasing speed. In order to appreciate variation of speed and velocity during projectile motion, we calculate the values of a projectile for successive seconds, which is projected with an initial velocity of 60 m/s making an angle of " 30 0 with the horizontal. Here, vertical component of velocity is 60 sin 30 0 " i.e. 30 m/s.


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Time      gt        Velocity      Magnitude of velocity
(s)     (m/s)         (m/s)            (m/s)
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0         0            30               30
1        10            20               20
2        20            10               10
3        30             0                0
4        40           -10               10
5        50           -20               20
6        60           -30               30
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Above table substantiate the observations made earlier. The magnitude of vertical velocity of the projectile first decreases during upward flight; becomes zero at maximum height; and, thereafter, picks up at the same rate during downward flight. It is also seen from the data that each of the magnitude of vertical velocity during upward motion is regained during downward motion. In terms of velocity, for every vertical velocity there is a corresponding vertical velocity of equal magnitude, but opposite in direction. The velocity – time plot of the motion is a straight line with negative slope. The negative slope here indicates that acceleration i.e acceleration due to gravity is directed in the opposite direction to that of positive y- direction.

Velocity – time plot The velocity – time plot for constant acceleration in vertical direction From the plot, we see that velocity is positive and acceleration is negative for upward journey, indicating deceleration i.e. decrease in speed. The time to reach maximum height in this case is : ⇒ t = u y g = 30 10 = 3 s The data in the table confirms this. Further, we know that vertical motion is independent of horizontal motion and time of flight for vertical motion is equal for upward and downward journey. This means that total time of flight is 2t i.e. 2x3 = 6 seconds. We must, however, be careful to emphasize that this result holds if the point of projection and point of return to the surface are on same horizontal level.

Projectile motion Initial and final velocities are equal in magnitude but opposite in direction. There is yet another interesting feature that can be drawn from the data set. The magnitude of vertical velocity of the projectile (30 m/s) at the time it hits the surface on return is equal to that at the time of the start of the motion. In terms of velocity, the final vertical velocity at the time of return is inverted initial velocity.

Displacement The displacement in vertical direction is given by : y = u y t - 1 2 g t 2

Projectile motion Vertical displacement at a given time This equation gives vertical position or displacement at a given time. It is important to realize that we have simplified the equation Δ y = y 2 - y 1 = u y t - 1 2 g t 2 by selecting origin to coincide by the point of projection so that, Δ y = y 2 - y 1 = y ( say ) Thus, “y” with this simplification represents position or displacement. The equation for position or displacement is a quadratic equation in time “t”. It means that solution of this equation yields two values of time for every value of vertical position or displacement. This interpretation is in fine agreement with the motion as projectile retraces all vertical displacement as shown in the figure.

Projectile motion All vertical displacement is achieved twice by the projectile except the point of maximum height. In case, initial and final points of the journey are on the same horizontal level, then the net displacement in vertical direction is zero i.e. y = 0. This condition allows us to determine the total time of flight “T” as : y = u y T - 1 2 g T 2 ⇒ 0 = u y T - 1 2 g T 2 ⇒ T ( u y - 1 2 g T ) = 0 ⇒ T = 0 or T = 2 u y g T = 0 corresponds to the time of projection. Hence neglecting the first value, the time of flight is : ⇒ T = 2 u y g We see that total time of flight is half of the time projectile takes to reach the maximum height. It means that projectile takes equal times in "up" and "down" motion. In other words, time of ascent equals time of descent. Problem : A ball is thrown upwards with a speed of 10 m/s making an angle 30° with horizontal and returning to ground on same horizontal level. Find (i) time of flight and (ii) and time to reach the maximum height Solution : Here, component of velocities in two directions are : u y = u sin θ = 10 sin 30 ° = 10 x 1 2 = 5 m / s (i) The time of flight, T, is : ⇒ T = 2 u y g = 2 x 5 10 = 1 s (ii) Time to reach the maximum height is half of the total flight when starting and end points of the projectile motion are at same horizontal level. Hence, the time to reach the maximum height is 0.5 s.

Equations of motion in horizontal direction The force due to gravity has no component in horizontal direction. This means that the motion in horizontal direction is not accelerated, but is an uniform motion. This implies that the component of velocity in x-direction is constant. As such, the position or displacement in x-direction at a given time “t” is : x = u x t

Projectile motion Horizontal displacement at a given time This equation gives the value of horizontal position or displacement at any given instant.

Displacement of projectile The displacement of projectile is obtained by vector addition of displacements in x and y direction. The magnitude of displacement of the projectile from the origin at any given instant is : Displacement, OP = ( x 2 + y 2 )

Displacement in projectile motion The angle that displacement vector subtends on x-axis is : tan α = y x

Velocity of projectile The velocity of projectile is obtained by vector addition of velocities in x and y direction. Since component velocities are mutually perpendicular to each other, we can find magnitude of velocity of the projectile at any given instant, applying Pythagoras theorem : v = ( v x 2 + v y 2 )

Velocity of a projectile The angle that the resultant velocity subtends on x-axis is : tan β = v y v x Problem : A ball is projected upwards with a velocity of 60 m/s at an angle 60° to the vertical. Find the velocity of the projectile after 1 second. Solution : In order to find velocity of the projectile, we need to know the velocity in vertical and horizontal direction. Now, initial velocities in the two directions are (Note that the angle of projection is given in relation to vertical direction.): u x = u sin θ = 60 sin 60 ° = 60 x 3 2 = 30 3 m / s u y = u cos θ = 60 cos 60 ° = 60 x 1 2 = 30 m / s Now, velocity in horizontal direction is constant as there is no component of acceleration in this direction. Hence, velocity after "1" second is : v x = u x = 30 3 m / s On the other hand, the velocity in vertical direction is obtained, using equation of motion as : v y = u y - g t ⇒ v y = 30 - 10 x 1 ⇒ v y = 20 m / s The resultant velocity, v, is given by : v = ( v x 2 + v y 2 ) ⇒ v = { ( 30 3 ) 2 + ( 20 ) 2 } = ( 900 x 3 + 400 ) = 55.68 m / s

Equation of the path of projectile Equation of projectile path is a relationship between “x” and “y”. The x and y – coordinates are given by equations, y = u y t - 1 2 g t 2 x = u x t Eliminating “t” from two equations, we have : y = u y x u x - g x 2 2 u x 2 For a given initial velocity and angle of projection, the equation reduces to the form of y = A x + B x 2 , where A and B are constants. The equation of “y” in “x” is the equation of parabola. Hence, path of the projectile motion is a parabola. Also, putting expressions for initial velocity components u x = u + cos θ and u y = u + sin θ , we have : ⇒ y = ( u sin θ ) x u cos θ - g x 2 2 u 2 cos 2 θ ⇒ y = x tan θ - g x 2 2 u 2 cos 2 θ