Time of flight, T

We have already determined the time of flight, which is given by :

This equation was derived in the earlier module Projectile motion with the assumption that both point of projection and point of return of the projectile lie on same horizontal level. It may be also be recalled that the equation of motion in vertical direction was evaluated for the condition that net displacement during the entire motion is zero. Hence, if the points are not on the same level, then above equation will not be valid and must be determined by equation of motion for the individual case with appropriate values.

From the above equation, we see that time of flight depends on initial speed and the angle of projection (θ). We must realize here that the range of θ is 0° < θ < 90°. For this range, sinθ is an increasing function. As such, we can say that a projection closer to vertical direction stays longer in the air for a given initial velocity. As a matter of fact, a verticle projectile for which θ = 90° and sinθ = 1, stays in the air for the maximum period.

Maximum height reached by the projectile, H

The vertical component velocity of the projectile reduces to zero as motion decelerates while going up against the force due to gravity. The point corresponding to this situation, when vertical component of velocity is zero, v y = 0 v y = 0 , is the maximum height that a projectile can reach. The projectile is accelerated downward under gravity immediately thereafter. Now, considering upward motion in y-direction,

Figure 1: Maximum height attained by projectile

Projectile motion

v y = u y - g t ⇒ 0 = u y - g t ⇒ t = u y g v y = u y - g t ⇒ 0 = u y - g t ⇒ t = u y g

Now, using equation for displacement in the vertical direction, we can find out the vertical displacement i.e. the height as :

y = u y t - 1 2 g t 2 ⇒ H = u y u y g - 1 2 u y g 2 ⇒ H = u y 2 g - u y 2 2 g ⇒ H = u y 2 2 g y = u y t - 1 2 g t 2 ⇒ H = u y u y g - 1 2 u y g 2 ⇒ H = u y 2 g - u y 2 2 g ⇒ H = u y 2 2 g

Putting expression of component velocity ( u y = u sin θ u y = u sin θ ), we have :

Just like the case of the time of flight, we see that maximum height reached by the projectile depends on both initial speed and the angle of projection (θ). Greater the initial velocity and greater the angle of projection from horizontal direction, greater is the height attained by the projectile.

It is important to realize here that there is no role of the horizontal component of initial velocity as far as maximum height is concerned. It is logical also. The height attained by the projectile is purely a vertical displacement; and as motions in the two mutually perpendicular directions are independent of each other, it follows that the maximum height attained by the projectile is completely determined by the vertical component of the projection velocity.

Range of projectile, R

The horizontal range is the displacement in horizontal direction. There is no acceleration involved in this direction. Motion is an uniform motion. It follows that horizontal range is transversed with the horizontal component of the projection velocity for the time of flight (T). Now,

Figure 2: The range of a projectile

Projectile motion

x = u x t ⇒ R = u x T = u cos θ x T x = u x t ⇒ R = u x T = u cos θ x T

where “T” is the time of flight. Putting expression for the time of flight,

⇒ R = u cos θ x 2 u sin θ g ⇒ R = u cos θ x 2 u sin θ g

Horizontal range, like time of flight and maximum height, is greater for greater projection speed.

Now, the range of the angle of projection with respect to horizontal direction, θ, is 0° < θ < 90° and the corresponding range of 2θ is 0° < 2θ < 180°. The “sin 2θ”, as appearing in the numerator for the expression of the horizontal range, is an increasing function for 0° < θ < 45° and a decreasing function for 45° < θ < 90° . For θ = 45°, sin 2θ = sin 90° = 1 (maximum).

In the nutshell, the range,R, increases with increasing angle of projection for 0° < θ < 45°; the range,R, is maximum when θ = 45°; the range,R, decreases with increasing angle of projection for 45° < θ < 90°.

The maximum horizontal range for a given projection velocity is obtained for θ = 45° as :

⇒ R = u 2 g ⇒ R = u 2 g

There is an interesting aspect of “sin 2θ” function that its value repeats for component angle i.e (90° – θ). For, any value of θ,

sin 2 ( 90 0 - θ ) = sin ( 180 0 - 2 θ ) = sin 2 θ sin 2 ( 90 0 - θ ) = sin ( 180 0 - 2 θ ) = sin 2 θ

Figure 3: Horizontal range is same for a pair of projection angle.

Horizontal range

It means that the range of the projectile with a given initial velocity is same for a pair of projection angles θ and 90° – θ. For example, if the range of the projectile with a given initial velocity is 30 m for an angle of projection, θ = 15°, then the range for angle of projection, θ = 90° - 15° = 75° is also 30m.

Impact of air resistance

We have so far neglected the effect of air resistance. It is imperative that if air resistance is significant then the features of a projectile motion like time of flight, maximum height and range are modified. As a matter of fact, this is the case in reality. The resulting motion is generally adversely affected as far as time of flight, maximum height and the range of the projectile are concerned.

Air resistance is equivalent to friction force for solid (projectile) and fluid (air) interface. Like friction, air resistance is self adjusting in certain ways. It adjusts to the relative speed of the projectile. Generally, greater the speed greater is air resistance. Air resistance also adjusts to the direction of motion such that its direction is opposite to the direction of relative velocity of two entities. In the nutshell, air resistance opposes motion and is equivalent to introducing a variable acceleration (resistance varies with the velocity in question) in the direction opposite to that of velocity.

For simplicity, if we consider that resistance is constant, then the vertical component of acceleration ( a y a y ) due to resistance acts in downward direction during upward motion and adds to the acceleration due to gravity. On the other hand, vertical component of air resistance acts in upward direction during downward motion and negates to the acceleration due to gravity. Whereas the horizontal component of acceleration due to air resistance ( a x a x )changes the otherwise uniform motion in horizontal direction to a decelerated motion.

Figure 4: Acceleration due to air resistance

Projectile motion with air resistance

With air resistance, the net or resultant acceleration in y direction depends on the direction of motion. During upward motion, the net or resultant vertical acceleration is " - g - a y - g - a y ". Evidently, greater vertical acceleration acting downward reduces speed of the particle at a greater rate. This, in turn, reduces maximum height. During downward motion, the net or resultant vertical acceleration is " - g + a y - g + a y ". Evidently, lesser vertical acceleration acting downward increases speed of the particle at a slower rate. On the other hand, the acceleration in x direction is " - a x - a x ". Clearly, the introduction of horizontal acceleration opposite to velocity reduces the range of the projectile (R).

What about time of flight ? Let us investigate. The projectile moves with slower speed and for a shorter trajectory. It is, therefore, clear that the time of flight decreases with air resistance.

Situations involving projectile motion

There are classic situations relating to the projectile motion, which needs to be handled with appropriate analysis. We have quite a few ways to deal with a particular situation. It is actually the nature of problem that would determine a specific approach from the following :

Besides, we may require combination of approached as listed above. In this section, we shall study these classic situations involving projectile motion.

Clearing posts of equal height

A projectile can clear posts of equal height, as projectile retraces vertical displacement attained during upward flight while going down. For this reason, the equation of motion for displacement yields two values for time for a given vertical displacement (height) : one corresponds to the time for upward flight and other for the downward flight as shown in the figure below.

Figure 5: The projectile retraces vertical displacement.

Projectile motion

Example 1

Problem : A projectile is thrown with a velocity of 25 2 25 2 m/s and at an angle 45° with the horizontal. The projectile just clears two posts of height 30 m each. Find (i) the position of throw on the ground from the post and (ii) find separation between the post.

Solution : Here, we first use the equation of displacement for the given height in the vertical direction to find the values of time when projectile clears the two posts. Once time instants are known, we can apply the equation of motion for uniform motion in horizontal direction to determine the distances as required.

Vertical motion :

Here,

u y = 25 2 x sin 45 0 = 25 m / s u y = 25 2 x sin 45 0 = 25 m / s

y = u y t - 1 2 g t 2 ⇒ 30 = 25 t - 1 2 25 x t 2 ⇒ t 2 - 5 t + 6 = 0 ⇒ t = 2 s or 3 s y = u y t - 1 2 g t 2 ⇒ 30 = 25 t - 1 2 25 x t 2 ⇒ t 2 - 5 t + 6 = 0 ⇒ t = 2 s or 3 s

Horizontal motion (refer the figure) :

OA = u x t = = 25 2 x cos 45 0 x 2 = 50 m OA = u x t = = 25 2 x cos 45 0 x 2 = 50 m

Thus, projectile needs to be thrown from a position 50 m from the pole. Now,

OB = u x t = = 25 2 x cos 45 0 x 3 = 50 m OB = u x t = = 25 2 x cos 45 0 x 3 = 50 m

Hence, separation, d, is :

d = OB - OA = 75 - 50 = 25 m d = OB - OA = 75 - 50 = 25 m

Hitting a specified target

An archer aims a bull’s eye; a person throws a pebble to strike an object placed at height and so on. The motion involved in these situations is a projectile motion – not a straight line motion. The motion of the projectile (arrow or pebble) has an arched trajectory due to gravity. We need to aim higher than line of sight to the object in order to negotiate the loss of height during flight.

Figure 6: The projectile hits the wall at a given point.

Projectile motion

As a particular height can be attained twice in the trajectory of projectile, the horizontal distance needs to be specified for the unique position of the target. Thus, this situation refers both horizontal and vertical displacements i.e x and y coordinates of the target with respect to the point of projection. For tihs reason, equation of path, that involves both x and y, suits the situation best.

y = x tan θ - g x 2 2 u 2 cos 2 θ y = x tan θ - g x 2 2 u 2 cos 2 θ

Example 2

Problem : A projectile, thrown at an angle 45° from the horizontal, strikes a building 30 m away at a point 15 above the ground. Find the velocity of projection.

Solution : As explained, the equation of projectile path suits the description of motion best. Here,

x = 30 m, y = 15 m and θ = 45°. Now,

y = x tan θ - g x 2 2 u 2 cos 2 θ y = x tan θ - g x 2 2 u 2 cos 2 θ

⇒ 15 = 30 tan 45 0 - 10 x 30 2 2 u 2 cos 2 45 0 ⇒ 15 = 30 x 1 - 10 x 2 x 30 2 2 u 2 ⇒ 15 = 30 - 9000 u 2 ⇒ u 2 = 600 ⇒ u = 24.49 m / s ⇒ 15 = 30 tan 45 0 - 10 x 30 2 2 u 2 cos 2 45 0 ⇒ 15 = 30 x 1 - 10 x 2 x 30 2 2 u 2 ⇒ 15 = 30 - 9000 u 2 ⇒ u 2 = 600 ⇒ u = 24.49 m / s

Check your understanding

Check the module titled " Features of projectile motion (Check your understanding) to test your understanding of the topics covered in this module.