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Projectile motion types

Module by: Sunil Kumar Singh. E-mail the author

Summary: A projectile may not return to the same level as the projection level. This difference of levels, however, does not change the basic approach. The motions in two mutually perpendicular directions are still independent of each other.

So far, we have limited our discussion to the classic mode of a projectile motion, where points of projection and return are on the same horizontal plane. This situation, however, may be altered. The projection level may be at an elevation with respect to the plane where projectile returns or the projection level may be at a lower level with respect to the plane where projectile returns. The two situations are illustrated in the figures below.

Figure 1
Projection and return at different levels
(a) Projection from higher elevation (b) The return of projectile at higher elevation
Figure 1(a) ( vpm1.gif)Figure 1(b) (vpm2.gif)

The two variants are basically the same parabolic motion. These motion types are inherently similar to the one where points of projection and return are on same level. If we look closely, then we find that the motion of the projectile from an elevation and from a lower level are either an extension or a curtailment of normal parabolic motion.

The projection on an incline (where point of return is on higher level) is a shortened projectile motion as if projectile has been stopped before returning to the normal point of return. This motion is also visualized as if the projectile is thrown over an incline or a wedge as shown in the figure. Again, there are two possibilities : (i) the projectile can be thrown up the incline or (ii) the projectile can be thrown down the incline. For the sake of convenience and better organization, we shall study projectile motion on an incline in a separate module. In this module, we shall restrict ourselves to the first case in which projectile is projected from an elevation.

Projection from a higher level

The projection of a projectile from a higher point results in a slightly different parabolic trajectory. We can visually recognize certain perceptible differences from the normal case as listed here :

Figure 2: The trajectory extends beyond the normal point of return.
Projection from higher level
Projection from higher level (vpm1.gif)

  • The upward trajectory is smaller than downward trajectory.
  • Time of ascent is smaller than the time of descent.
  • The speed of projection is not equal to speed of return on the ground.
  • The velocity of return is more aligned to vertical as the motion progresses.

It is evident that the expressions derived earlier for time of flight (T), maximum height (H) and range (R) are not valid in the changed scenario. But the basic consideration of the analysis is necessarily same. The important aspect of projectile motion that motions in two mutually perpendicular directions are independent of each other, still, holds. Further, the nature of motion in two directions is same as before : the motion in vertical direction is accelerated due to gravity, whereas motion in horizontal direction has no acceleration.

Now, there are two important variations of this projectile motion, when projected from an elevated level. The projectile may either be projected at certain angle (up or down) with the horizontal or it may be projected in the horizontal direction.

Figure 3
The projection from higher elevation
(a) Projection from higher elevation at an angle (b) Projection from higher elevation in horizontal direction
Figure 3(a) ( vpm1.gif)Figure 3(b) (vpm6.gif)

There are many real time situations that resemble horizontal projection. When an object is dropped from a plane flying parallel to the ground at certain height, then the object acquires horizontal velocity of the plane when the object is released. As the object is simply dropped, the velocity in vertical direction is zero. This horizontal velocity of the object, as acquired from the plane, is then modified by the force of gravity, whereby the object follows a parabolic trajectory before hitting the ground.

This situation is analogous to projection from ground except that we track motion from the highest point. Note that vertical velocity is zero and horizontal velocity is tangential to the path at the time of projection. This is exactly the same situation as when projectile is projected from the ground and reaches highest point. In the nutshell, the description of motion here is same as the description during descent when projected from the ground.

Figure 4: The position of plane is above object as both moves with same velocity in horizontal direction.
An object dropped from a plane moving in horizontal direction
An object dropped from a plane moving in horizontal direction (vpm4.gif)

The interesting aspect of the object dropped from plane is that both plane and object are moving with same horizontal velocity. Hence, plane is always above the dropped object, provided plane maintains its velocity.

The case of projection from a higher level at certain angle (up or down) to the horizontal is different to the one in which projectile is projected horizontally. The projectile has a vertical component of initial velocity when thrown at an angle with horizontal. This introduces the difference between two cases. The projectile thrown up attains a maximum height above the projection level. On the return journey downward, it travels past its level of projection. The difference is visually shown in the two adjoining figures below.

Figure 5
Maximum height attained by the projectile
(a) (b)
Figure 5(a) ( vpm5.gif)Figure 5(b) (vpm7.gif)

The resulting trajectory in the first case has both upward and downward motions. On the other hand, the motion in upward direction is completely missing in the horizontal projection as the projectile keeps loosing altitude all the time.

Projectile thrown in horizontal direction

We can easily analyze projectile motion following the technique of component motions in two mutually perpendicular directions (horizontal and vertical). Typically, we consider vertical component of motion to determine time of flight (T). The initial velocity in vertical direction is zero.

We consider point of projection as origin of coordinate system. Further, we choose x-axis in horizontal direction and y-axis in the vertically downward direction for the convenience of analysis. Then,

Figure 6
An object projected in horizontal direction
An object projected in horizontal direction (vpm10.gif)

y = H = u y T + 1 2 g T 2 y = H = u y T + 1 2 g T 2

But u y = 0 u y = 0 ,

⇒ H = 1 2 g T 2 ⇒ H = 1 2 g T 2

⇒ T = 2 H g ⇒ T = 2 H g

Note the striking similarity here with the free fall of a body under gravity from a height “H”. The time taken in free fall is same as the time of flight of projectile in this case. Now, the horizontal range of the projectile is given as :

x = R = u x T = u 2 H g x = R = u x T = u 2 H g

Example 1

Problem : A plane flying at the speed of 100 m/s parallel to the ground drops an object from a height of 2 km. Find (i) the time of flight (ii) velocity of the object at the time it strikes the ground and (iii) the horizontal distance traveled by the object.

Solution : The basic approach to solve the problem involves consideration of motion in two mutually perpendicular direction. Here, we consider a coordinate system with the point of release as the origin and down ward direction as the positive y-direction.

Figure 7
An object dropped from a plane moving in horizontal direction
An object dropped from a plane moving in horizontal direction (vpm8a.gif)

(i) Time of flight, T

In vertical direction :

u y = 0 , a = 10 m / s 2 , y = 2000 m , T = ? u y = 0 , a = 10 m / s 2 , y = 2000 m , T = ?

Using equation, y = u y T + 1 2 g T 2 y = u y T + 1 2 g T 2 , we have :

⇒ y = 1 2 g t 2 ⇒ T = ( 2 y g ) ⇒ T = ( 2 x 2000 10 ) = 20 s ⇒ y = 1 2 g t 2 ⇒ T = ( 2 y g ) ⇒ T = ( 2 x 2000 10 ) = 20 s

(ii) Velocity at the ground

We can find the velocity at the time of strike with ground by calculating component velocities at that instant in the two mutually perpendicular directions and finding the resultant (composite) velocity as :

v = ( v x 2 + v y 2 ) v = ( v x 2 + v y 2 )

Initial vertical component of initial velocity (uy) is zero and the object is accelerated down with the acceleration due to gravity. Hence,

v y = u y + g t = 0 + g t = g t v y = u y + g t = 0 + g t = g t

The component of velocity in the horizontal direction remains unchanged as there is no acceleration in this direction.

v x = u x = u v x = u x = u

⇒ v = ( v x 2 + v y 2 ) = ( u 2 + g 2 t 2 ) ⇒ v = ( v x 2 + v y 2 ) = ( u 2 + g 2 t 2 )

Putting values,

⇒ v = ( 100 2 + 10 2 x 20 2 ) = 50000 = 100 5 m / s ⇒ v = ( 100 2 + 10 2 x 20 2 ) = 50000 = 100 5 m / s

(iii) Horizontal distance traveled

From consideration of uniform motion in horizontal direction, we have :

x = u x t = u t x = u x t = u t

Putting values,

⇒ x = R = 100 x 20 = 2000 m ⇒ x = R = 100 x 20 = 2000 m

Exercise 1

A ball is thrown horizontally from a tower at a speed of 40 m/s. The speed of the projectile (in m/s) after 3 seconds, before it touches the ground, is (consider g = 10 m / s 2 m / s 2 ) :

a 30 b 40 c 50 d 60 a 30 b 40 c 50 d 60

Solution

Here, we consider a reference system whose origin coincides with the point of projection. The downward direction is along y - direction as shown in the figure.

Figure 8: Projectile motion
Projectile motion
Projectile motion (pq1.gif)

The initial speed of the projectile is equal to the horizontal component of the velocity, which remains unaltered during projectile motion. On the other hand, vertical component of velocity at the start of motion is zero. Thus,

u x = 40 m / s , u y = 0 u x = 40 m / s , u y = 0

Using equation of motion, we have :

v y = u y + a y t v y = u y + a y t ⇒ v y = 0 + g t = 10 X 3 = 30 m / s ⇒ v y = 0 + g t = 10 X 3 = 30 m / s

Since the horizontal component of velocity remains unaltered, the speed, after 3 second, is :

⇒ v = v x 2 + v y 2 = 40 2 + 30 2 = 50 m / s ⇒ v = v x 2 + v y 2 = 40 2 + 30 2 = 50 m / s Hence, option (c) is correct.

Exercise 2

A ball is projected horizontally from a height at a speed of 30 m/s. The time after which the vertical component of velocity becomes equal to horizontal component of velocity is : (consider g = 10 m / s 2 m / s 2 ) :

a 1 s b 2 s c 3 s d 4 s a 1 s b 2 s c 3 s d 4 s

Solution

The ball does not have vertical component of velocity when projected. The ball, however, is accelerated downward and gains speed in vertical direction. At certain point of time, the vertical component of velocity equals horizontal component of velocity. At this instant, the angle that the velocity makes with the horizontal is :

Figure 9: Projectile motion
Projectile motion
Projectile motion (pq1.gif)

tan θ = v y v x = 1 tan θ = v y v x = 1 ⇒ θ = 45 0 ⇒ θ = 45 0

We should note that this particular angle of 45° at any point during the motion, as a matter of fact, signifies that two mutually perpendicular components are equal.

But we know that horizontal component of velocity does not change during the motion. It means that vertical component of velocity at this instant is equal to horizontal component of velocity i.e.

v y = v x = u x = 30 m / s v y = v x = u x = 30 m / s

Further, we know that we need to analyze motion in vertical direction to find time as required,

v y = u y + a y t v y = u y + a y t ⇒ 30 = 0 + 10 t ⇒ 30 = 0 + 10 t ⇒ t = 3 s ⇒ t = 3 s

Hence, option (c) is correct.

Projectile thrown at an angle with horizontal direction

There are two possibilities. The projectile can be projected up or down as shown in the figure here :

Figure 10
Projection from an elevated level
(a) Projection upwards from an elevated level (b) Projection downwards from an elevated level
Figure 10(a) (vpm11a.gif)Figure 10(b) (vpm12a.gif)

Projectile thrown up at an angle with horizontal direction

The time of flight is determined by analyzing motion in vertical direction. The net displacement during the motion is equal to the elevation of point of projection above ground i.e. H 2 H 2 . To analyze the motion, we consider point of projection as origin, horizontal direction as x-axis and upward vertical direction as y-axis.

y = - H 2 = u y T − 1 2 g T 2 y = - H 2 = u y T − 1 2 g T 2

Rearranging,

T 2 − 2 u y g T + 2 H 2 g = 0 T 2 − 2 u y g T + 2 H 2 g = 0

This is a quadratic equation in “T”. Solving we get two values of T, one of which gives the time of flight.

Figure 11
An object projected up at an angle from an elevation
An object projected up at an angle from an elevation (vpm11.gif)

Horizontal range is given by analyzing motion in horizontal direction as :

x = R = u x T x = R = u x T

While calculating maximum height, we can consider motion in two parts. The first part is the motion above the projection level. On the other hand second part is the projectile motion below projection level. The total height is equal to the sum of the magnitudes of vertical displacements :

H = | H 1 | + | H 2 | H = | H 1 | + | H 2 |

Example 2

Problem : A projectile is thrown from the top of a building 160 m high, at an angle of 30° with the horizontal at a speed of 40 m/s. Find (i) time of flight (ii) Horizontal distance covered at the end of journey and (iii) the maximum height of the projectile above the ground.

Solution : Unlike horizontal projection, the projectile has a vertical component of initial velocity. This vertical component is acting upwards, which causes the projectile to rise above the point of projection.

Here, we choose the point of projection as the origin and downward direction as the positive y – direction.

Figure 12
An object projected at an angle from an elevation
An object projected at an angle from an elevation (vpm9a.gif)

(i) Time of flight, T

Here,

u y = u sin θ = - 40 sin 30 0 = - 20 m / s ; y = 160 m ; g = 10 m / s 2 u y = u sin θ = - 40 sin 30 0 = - 20 m / s ; y = 160 m ; g = 10 m / s 2

Using equation, y = u y t + 1 2 g t 2 y = u y t + 1 2 g t 2 , we have :

⇒ 160 = - 20 t + 1 2 10 t 2 ⇒ 5 t 2 - 20 t - 160 = 0 ⇒ t 2 - 4 t - 32 = 0 ⇒ t 2 - 8 t + 4 t - 32 = 0 ⇒ t ( t - 8 ) + 4 ( t - 8 ) = 0 ⇒ t = - 4 s or t = 8 s ⇒ 160 = - 20 t + 1 2 10 t 2 ⇒ 5 t 2 - 20 t - 160 = 0 ⇒ t 2 - 4 t - 32 = 0 ⇒ t 2 - 8 t + 4 t - 32 = 0 ⇒ t ( t - 8 ) + 4 ( t - 8 ) = 0 ⇒ t = - 4 s or t = 8 s

Neglecting negative value of time, T = 8 s.

(ii) Horizontal distance, R

There is no acceleration in horizontal direction. Using equation for uniform motion,

x = u x T x = u x T

Here,

u x = u cos θ = 40 cos 30 0 = 20 3 m / s ; T = 8 s u x = u cos θ = 40 cos 30 0 = 20 3 m / s ; T = 8 s

⇒ x = u x T = 20 3 x 8 = 160 3 m ⇒ x = u x T = 20 3 x 8 = 160 3 m

(iii) Maximum height, H

The maximum height is the sum of the height of the building ( H 2 H 2 ) and the height attained by the projectile above the building ( H 1 H 1 ).

H = H 1 + H 2 H = H 1 + H 2

We consider vertical motion to find the height attained by the projectile above the building ( H 1 H 1 ).

H 1 = u 2 sin 2 θ 2 g H 1 = u 2 sin 2 θ 2 g

⇒ H 1 = 40 2 sin 2 30 0 2 X 10 ⇒ H 1 = 40 2 sin 2 30 0 2 X 10

⇒ H 1 = 40 2 X 1 4 2 X 10 = 20 m ⇒ H 1 = 40 2 X 1 4 2 X 10 = 20 m

Thus maximum height, H, is :

H = H 1 + H 2 = 20 + 160 = 180 m H = H 1 + H 2 = 20 + 160 = 180 m

Projectile thrown down at an angle with horizontal direction

Projectile motion here is similar to that of projectile thrown horizontally. The only difference is that the projectile has a finite component of velocity in downward direction against zero vertical velocity.

For convenience, the point of projection is considered as origin of reference and the positive x and y directions of the coordinate system are considered in horizontal and vertically downward directions.

Figure 13
An object projected down at an angle from an elevation
An object projected down at an angle from an elevation (vpm12.gif)

The time of flight is obtained considering motion in vertical direction as :

y = H = u y T + 1 2 g T 2 y = H = u y T + 1 2 g T 2

Rearranging, we get a quadratic equation in “T”,

T 2 + 2 u y g T − 2 H 1 g = 0 T 2 + 2 u y g T − 2 H 1 g = 0

One of the two values of "T" gives the time of flight in this case. The horizontal range, on the other hand, is given as :

x = R = u x T x = R = u x T

Exercises

Exercise 3

Two balls of masses " m 1 m 1 ” and “ m 2 m 2 ”are thrown from a tower in the horizontal direction at speeds " u 1 u 1 ” and “ u 2 u 2 ” respectively at the same time. Which of the two balls strikes the ground first?

(a) the ball thrown with greater speed

(b) the ball thrown with lesser speed

(c) the ball with greater mass

(d) the balls strike the ground simultaneously

Solution

The time to strike the ground is obtained by considering motion in vertical direction.

Here, u y u y = 0 and y=T (total time of flight)

y = 1 2 g T 2 y = 1 2 g T 2 ⇒ T = 2 y g ⇒ T = 2 y g

Thus, we see that time of flight is independent of both mass and speed of the projectile in the horizontal direction. The two balls, therefore, strike the ground simultaneously.

Hence, option (d) is correct.

Exercise 4

A body dropped from a height "h" strikes the ground with a velocity 3 m/s. Another body of same mass is projected horizontally from the same height with an initial speed of 4 m/s. The final velocity of the second body (in m/s), when it strikes the earth will be :

a 3 b 4 c 5 d 7 a 3 b 4 c 5 d 7

Solution

We can use the fact that velocities in vertical direction, attained by two bodies through same displacement, are equal. As such, velocity of the second body , on reaching the ground, would also be 3 m/s. Now, there is no acceleration in horizontal direction. The horizontal component of velocity of the second body, therefore, remains constant.

v x = 4 m / s v x = 4 m / s

and

v y = 3 m / s v y = 3 m / s

The resultant velocity of two component velocities in mutually perpendicular directions is :

⇒ v = v x 2 + v y 2 = 4 2 + 3 2 = 5 m / s ⇒ v = v x 2 + v y 2 = 4 2 + 3 2 = 5 m / s

Exercise 5

A projectile (A) is dropped from a height and another projectile (B) is projected in horizontal direction with a speed of 5 m/s from the same height. Then the correct statement(s) is(are) :

(a) The projectile (A) reach the ground earlier than projectile (B).

(b) Both projectiles reach the ground with the same speed.

(c) Both projectiles reach the ground simultaneously.

(d) The projectiles reach the ground with different speeds.

Solution

First three options pertain to the time of flight. The time of flight depends only the vertical displacement and vertical component of projection velocity. The vertical components of projection in both cases are zero. The time of flight (T) in either case is obtained by considering motion in vertical direction as :

y = u y T + 1 2 a y T 2 y = u y T + 1 2 a y T 2 ⇒ H = 0 + 1 2 g T 2 ⇒ H = 0 + 1 2 g T 2 ⇒ T = 2 H g ⇒ T = 2 H g

Hence, both projectiles reach the ground simultaneously. On the other hand, component of velocity in vertical direction, on reaching the ground is :

v y = u y + a y v y = u y + a y ⇒ T = 0 + g T ⇒ T = 0 + g T

Putting value of T, we have :

⇒ v y = 2 g H ⇒ v y = 2 g H

Projectile (A) has no component of velocity in horizontal direction, whereas projectile (B) has finite component of velocity in horizontal direction. As such, velocities of projectiles and hence the speeds of the particles on reaching the ground are different.

Hence, options (c) and (d) are correct.

Exercise 6

Four projectiles, "A","B","C" and "D" are projected from top of a tower with velocities (in m/s) 10 i + 10 j, 10 i - 20 j , -10 i - 10 j and -20 i + 10j in the coordinate system having point of projection as origin. If "x" and "y" coordinates are in horizontal and vertical directions respectively, then :

(a)Time of flight of C is least.

(b) Time of flight of B is least.

(c) Times of flights of A and D are greatest.

(d) Times of flights of A and D are least.

Solution

The time of flight of a projectile solely depends on vertical component of velocity.

Figure 14: Projectile motion
Projectile motion
Projectile motion (pq3.gif)

The projectile "A" is thrown up from an elevated point with vertical component of velocity 10 m/s. It travels to the maximum height ( H 1 H 1 ) and the elevation from the ground( H 2 H 2 ).

The projectile "B" is thrown down from an elevated point with vertical component of velocity 20 m/s. It travels only the elevation from the ground ( H 2 H 2 ).

The projectile "C" is thrown down from an elevated point with vertical component of velocity 10 m/s. It travels only the elevation from the ground ( H 2 H 2 ).

The projectile "D" is thrown up from an elevated point with vertical component of velocity 15 m/s. It travels to the maximum height ( H 1 H 1 ) and the elevation from the ground( H 2 H 2 ).

Clearly, projectile "B" and "C" travel the minimum vertical displacement. As vertical downward component of velocity of "B" is greater than that of "C", the projectile "B" takes the least time.

The projectiles "A" and "D" are projected up with same vertical components of velocities i.e. 10 m/s and take same time to travel to reach the ground. As the projectiles are projected up, they take more time than projectiles projected down.

Hence, options (b) and (c) are correct.

Acknowledgment

Author wishes to thank Scott Kravitz, Programming Assistant, Connexions for making suggestion to remove syntax error in the module.

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