Projectiles

In this tutorial, you will learn the following:

  • Projectiles
  • Vertical and horizontal motion of projectiles
  • Three different scenarios of projectiles with worked examples
  • How to maximize the horizontal range of a projectile

Projectile

An object that moves under gravity in the air is a projectile.

E.g.

  • › A cricket ball that moves through the air, after being struck by a bat
  • › A baseball that moves through the air, after being struck by a baseball bat
  • › A football that moves in the air after being kicked
  • › A dart that moves through the air
  • › An arrow that moves through the air, having been fired by an archer

Suppose an object is projected at an angle of t to the horizontal at the speed of v. Since it moves both vertically and horizontally, the initial velocity has components both vertically and horizontally: the vertical component takes it up and the horizontal component takes it horizontally. The following image illustrates it:

projectile motion - horizontal and vertical components

Since there is no horizontal force at work in the absence of air resistance, according to F = ma, the acceleration is zero. So, the horizontal velocity remains constant throughout the motion. The vertical velocity, however, goes down, as the gravitational pull acts in the opposite direction - downwards. When the projectile reaches the peak, it becomes zero and then changes the direction and increases as it goes down.

You can practise the changes in velocity of a projectile interactively with the following applet; just move the slider - time - and see the changes:

Maximum Height
This is he maximum vertical height reached by the projectile.

Time of flight
This is the time during which the object has been in the air.

Maximum Horizontal Range
This is he maximum horizontal distance travelled by the projectile.

The following worked examples cover the above in detail.

Projectiles - Problem Solving

E.g.1

A ball is projected at 20√2 m/s, at an angle of 450 to the horizontal. Find the following, by assuming that g = 10ms-2:

  1. The maximum height
  2. The time of flight
  3. The maximum horizontal range
projectile motion - problem 1
  1. For vertical Motion, until the objects comes to an instant halt
    ↑ v2 = u2 + 2as
    u = v sin(45) = 20√2 &1/√2 = 20; a = -10; s = h
    0 = 400 - 2x10xs
    h = 20m.
  2. When the object reaches the highest point,
    ↑ v = u + at
    u = v sin(45) = 20√2 &1/√2 = 20; a = -10; v = 0
    0 = 20 - 10t
    t = 2s
    The time of flight = 2x2 = 4s.
  3. → s = ut + 1/2 at2 u = 20√2 cos(45) = 20 x 1/√2 = 20; a = 0; t = 4
    s = 20x4 + 0
    s = 80m.

E.g.2

A ball is projected at 20 m/s, horizontally from a tower of height 20m. Find the following, by assuming that g = 10ms-2:

  1. The time of flight
  2. The maximum horizontal range
  3. The velocity after 1 second
  4. The velocity at which it hits the ground
projectile motion - problem 2

Since it was projected horizontally, there would be no vertical speed at the beginning.

  1. For vertical Motion, until it hits the ground,
    ↓ s = ut + 1/2 at2
    u = 0; a = 10; s = 20
    20 = 0 + 5t2
    t2 = 4 => t = ±2
    t = 2s.
  2. When the object reaches the ground,
    → s = ut + 1/2 at2
    u = 20; a = 0; t = 2
    s = 20x2 + 0
    s = 40m
  3. After 1s, it stats moving downward; it has a vertical velocity and a horizontal velocity, which is constant.
    ↓ v = u + at
    u = 0; a = 10; t = 1
    v = 0 + 1x10
    v = 10ms-1.
    → v = u + at
    u = 20; a = 0; t = 1
    v = 20 + 0
    v = 20ms-1.
    So, the resultant velocity = √(102 + 202)
    ↘ v = √500 = 10√5 m/s.
  4. When it hits the ground too, it has a vertical velocity and a horizontal velocity, which is constant.
    ↓ v = u + at
    u = 0; a = 10; t = 2
    v = 0 + 2x10
    v = 20ms-1.
    → v = u + at
    u = 20; a = 0; t = 1
    v = 20 + 0
    v = 20ms-1.
    So, the resultant velocity = √(202 + 202)
    ↘ v = √800 = 20√2 m/s.

E.g.3

A ball is projected at 20√2, at an angle of 450 to the horizontal, from the top of a tower of height 60m, upwards. Find the following, by assuming that g = 10ms-2:

  1. The time of flight
  2. The maximum horizontal range
  3. The maximum height reached by the object
  4. The velocity at which it hits the ground
projectile motion - problem 3

When the object reaches the level of the top of the tower again, its displacement is zero; below that level, it is negative.

  1. For vertical Motion, until it reaches the ground,
    ↑ s = ut + 1/2 at2
    u = 20√2 sin(45) = 20; a = -10; s = -60
    -60 = 20 - 5t2
    t2 = 16 => t = ±4
    t = 4s.
  2. When the object reaches the ground,
    → s = ut + 1/2 at2
    u = 20√2 cos(45) = 20; a = 0; t = 4
    s = 20x4 + 0
    s = 80m
  3. ↑ v2 = u2 - 2as
    u = 20; a = -10; v = 0
    0 = 400 - 2x10xs
    s = 20m
    Total height = 20 + 60 = 80m.
  4. ↓ v = u + at
    u = 20; a = -10; v = ? t = 4
    v = 20 - 10x4 = -20m/s
    → since horizontal velocity remains the same, v = 20
    So, the resultant velocity = √(202 + 202)
    ↘ v = √800 = 20√2 m/s.

Maximizing the Horizontal Range

Suppose an object is projected at an angle, t, to the horizontal at v.

projectile motion - maximum horizontal range

When it reaches the peak,
↑ u = v sin(t); v = 0; t = t
o = v sin(t) - at
t = v sin(t)/a
The time of flight= 2v sin(t)/a
→ s = x; u = v cos(t); a = 0; t = 2vsin(t)
x = 2v2sin(t)cos(t)
Since 2 sin(t) cos(t) = sin(2t) - a trigonometric identity
x = v2sin(2t)
dx/dt = 2v2cos(2t)
When x is maximum, dx/dt =0
2v2cos(2t) = 0
cos(2t) = 0
2t = 900
2t = 450
So, the horizontal range is maximum when the angle of projection from the horizontal is 450.
That means, if a cricketer or a footballer think of maximizing the horizontal range, here is the way!

 

 

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