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Projectile motion types (application)

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

Summary: Solving problems is an essential part of the understanding process.

Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

Hints on solving problems

1: Identify projectile motion types. The possible variants are :

  • Projectile is thrown in horizontal direction. In this case, initial vertical component of velocity is zero. Consider horizontal direction as positive x-direction and vertically downward direction as positive y-direction.
  • Projectile is thrown above horizontal level. The projectile first goes up and then comes down below the level of projection
  • Projectile is thrown below horizontal level. Consider horizontal direction as positive x-direction and vertically downward direction as positive y-direction.

2: We can not use standard equations of time of flight, maximum height and horizontal range. We need to analyze the problem in vertical direction for time of flight and maximum height. Remember that determination of horizontal range will involve analysis in both vertical (for time of flight) and horizontal (for the horizontal range) directions.

3: However, if problem has information about motion in horizontal direction, then it is always advantageous to analyze motion in horizontal direction.

Representative problems and their solutions

We discuss problems, which highlight certain aspects of the study leading to the projectile motion types. The questions are categorized in terms of the characterizing features of the subject matter :

  • Time of flight
  • Range of flight
  • Initial velocity
  • Final velocity

Time of flight

Example 1

Problem : A ball from a tower of height 30 m is projected down at an angle of 30° from the horizontal with a speed of 10 m/s. How long does ball take to reach the ground? (consider g = 10 m / s 2 m / s 2 )

Solution : Here, we consider a reference system whose origin coincides with the point of projection. Further, we consider that the downward direction is positive y - direction.

Figure 1
Projectile motion
 Projectile motion  (pmtq1ab.gif)

Motion in vertical direction :

Here, u y = u sin θ = 10 sin 30 0 = 5 m / s ; y = 30 m ; u y = u sin θ = 10 sin 30 0 = 5 m / s ; y = 30 m ; . Using y = u y t + 1 2 a y t 2 y = u y t + 1 2 a y t 2 , we have :

30 = 5 t + 1 2 10 t 2 t 2 + t - 6 = 0 t ( t + 3 ) - 2 ( t + 3 ) = 0 t = - 3 s or t = 2 s 30 = 5 t + 1 2 10 t 2 t 2 + t - 6 = 0 t ( t + 3 ) - 2 ( t + 3 ) = 0 t = - 3 s or t = 2 s

Neglecting negative value of time, t = 2 s

Range of flight

Example 2

Problem : A ball is thrown from a tower of height “h” in the horizontal direction at a speed “u”. Find the horizontal range of the projectile.

Solution : Here, we consider a reference system whose origin coincides with the point of projection. we consider that the downward direction is positive y - direction.

Figure 2
Projectile motion
 Projectile motion  (pmtq5a.gif)

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

Motion in the vertical direction :

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

h = 1 2 g T 2 T = ( 2 h g ) h = 1 2 g T 2 T = ( 2 h g )

Putting expression of total time of flight in the expression for horizontal range, we have :

R = u ( 2 h g ) R = u ( 2 h g )

Example 3

Problem : A projectile is projected up with a velocity √(2ag) at an angle “θ” from an elevated position as shown in the figure. Find the maximum horizontal range that can be achieved.

Figure 3: Projectile projected from an elevated point .
Projectile motion
 Projectile motion   (pmtq6a.gif)

Solution : In order to determine the maximum horizontal range, we need to find an expression involving horizontal range. We shall use the equation of projectile as we have the final coordinates of the motion as shown in the figure below :

Figure 4: Projectile projected from an elevated point .
Projectile motion
 Projectile motion   (pmtq7a.gif)

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

Substituting and changing trigonometric ratio with the objective to create a quadratic equation in “tan θ” :

H = R tan θ g R 2 2 { 2 a g } 2 1 + tan 2 θ H = R tan θ g R 2 2 { 2 a g } 2 1 + tan 2 θ

Rearranging, we have :

R 2 tan 2 θ 4 a R tan θ + R 2 4 a H = 0 R 2 tan 2 θ 4 a R tan θ + R 2 4 a H = 0

tan θ = 4 a R ± { 4 a R 2 4 R 2 R 2 4 a H } 2 R 2 tan θ = 4 a R ± { 4 a R 2 4 R 2 R 2 4 a H } 2 R 2

For tan θ to be real, it is required that

16 a 2 R 2 4 R 2 R 2 4 a H 16 a 2 R 2 4 R 2 R 2 4 a H

4 a 2 R 2 4 a H 4 a 2 R 2 4 a H

R 2 4 a a + H R 2 4 a a + H

R ± 2 a a + H R ± 2 a a + H

Hence, maximum possible range is :

R = 2 a a + H R = 2 a a + H

Initial velocity

Example 4

Problem : A ball is thrown horizontally from the top of the tower to hit the ground at an angle of 45° in 2 s. Find the speed of the ball with which it was projected.

Solution : The question provides the angle at which the ball hits the ground. A hit at 45° means that horizontal and vertical speeds are equal.

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

v x = v y v x = v y

However, we know that horizontal component of velocity does not change with time. Hence, final velocity in horizontal direction is same as initial velocity in that direction.

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

We can now find the vertical component of velocity at the time projectile hits the ground by considering motion in vertical direction. Here, u y = 0, t = 2 s . u y = 0, t = 2 s .

Using equation of motion in vertical direction, assuming downward direction as positive :

v y = u y + a t v y = u y + a t

v y = 0 + 10 X 2 = 20 m / s v y = 0 + 10 X 2 = 20 m / s

Hence, the speed with which the ball was projected in horizontal direction is :

u x = v y = 20 m / s u x = v y = 20 m / s

Final velocity

Example 5

Problem : A ball “A” is thrown from the edge of building “h”, at an angle of 30° from the horizontal, in upward direction. Another ball ”B” is thrown at the same speed from the same position, making same angle with horizontal, in vertically downward direction. If "u" be the speed of projection, then find their speed at the time of striking the ground.

Solution : The horizontal components of velocity for two projectiles are equal. Further, horizontal component of velocity remains unaltered during projectile motion. The speed of the projectile at the time of striking depends solely on vertical component of velocity. For the shake of convenience of analysis, we consider point of projection as the origin of coordinate system and vertically downward direction as positive y - direction.

Figure 5
Projectile motion
 Projectile motion  (pmtq2a.gif)

Motion in vertical direction :

For A, u yA = - u sin 30 0 = - u 2 u yA = - u sin 30 0 = - u 2

For B, u yB = u sin 30 0 = u 2 u yB = u sin 30 0 = u 2

Thus, velocities in vertical direction are equal in magnitude, but opposite in direction. The ball,"A", which is thrown upward, returns after reaching the maximum vertical height. For consideration in vertical direction, the ball returns to the point of projection with same speed it was projected. What it means that the vertical component of velocity of ball "A" on return at the point projection is "u/2". This further means that two balls "A" and "B", as a matter of fact, travel down with same downward vertical component of velocity.

In other words, the ball "A" returns to its initial position acquiring same speed "u/2" as that of ball "B" before starting its downward journey. Thus, speeds of two balls are same i.e "u/2" for downward motion. Hence, two balls strike the ground with same speed. Let the final speed is "v".

Now, we apply equation of motion to determine the final speed in the vertical direction.

v y 2 = u y 2 + 2 g h v y 2 = u y 2 + 2 g h

Putting values, we have :

v y 2 = u 2 2 + 2 g h = u 2 + 8 g h 4 v y 2 = u 2 2 + 2 g h = u 2 + 8 g h 4

v y = u 2 + 8 g h 2 v y = u 2 + 8 g h 2

The horizontal component of velocity remains same during the journey. It is given as :

v x = u cos 30 0 = 3 u 2 v x = u cos 30 0 = 3 u 2

The resultant of two mutually perpendicular components is obtained, using Pythagoras theorem :

v 2 = v x 2 + v y 2 = 3 u 2 4 + u 2 + 8 g h 4 v 2 = v x 2 + v y 2 = 3 u 2 4 + u 2 + 8 g h 4

v 2 = 3 u 2 + u 2 + 8 g h 4 = u 2 + 2 g h v 2 = 3 u 2 + u 2 + 8 g h 4 = u 2 + 2 g h

v = u 2 + 2 g h v = u 2 + 2 g h

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