Summary: The motion of the projectile has an arched trajectory due to gravity.
The span of projectile motion in the vertical plane is determined by two factors, namely the speed of projection and angle of projection with respect to horizontal. These two factors together determine (i) how long does the projectile remain in air (time of flight, T) (ii) how far does the projectile go in the horizontal direction (range of projectile, R) and (iii) how high does the projectile reach (maximum height, H).
Further, the trajectory of the projectile is symmetric about a vertical line passing through the point of maximum height if point of projection and point of return fall on the same horizontal surface.
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.
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,
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Now, using equation for displacement in the vertical direction, we can find out the vertical displacement i.e. the height as :
Putting expression of component velocity (
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.
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,
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where “T” is the time of flight. Putting expression for the time of flight,
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 :
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,
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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.
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 (
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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
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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.
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.
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.
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Problem : A projectile is thrown with a velocity of
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,
Horizontal motion (refer the figure) :
Thus, projectile needs to be thrown from a position 50 m from the pole. Now,
Hence, separation, d, is :
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.
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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.
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,
A projectile trajectory under gravity is completely determined by the initial speed and the angle of projection or simply by the initial velocity (direction is implied). For the given velocity, maximum height and the range are unique – notably independent of the mass of the projectile.
Thus, a projectile motion involving attributes such as maximum height and range is better addressed in terms of the equations obtained for the specific attributes of the projectile motion.
Problem : Determine the angle of projection for which maximum height is equal to the range of the projectile.
Solution : We equate the expressions of maximum height and range (H = R) as :
Check the module titled " Features of projectile motion (Check your understanding) to test your understanding of the topics covered in this module.