Projectile Motion Notes

Dynamics of Projectile Motion

• Horizontal and vertical velocity components

• Projectile motion calculation

• Relationship between range and launched angle

• The effect of air resistance

Revision: Equations of Linear Motion

• For an object moving with uniform velocity: $s = vt$

• For an object moving with uniform acceleration:

  • $s=ut+\frac12at^2$

  • $v = u + at$

  • $v^2 = u^2 + 2as$

• LINEAR MOTION = motion along a straight line

• FREE FALL = a body is said to experience free fall if it falls under action of gravity in vacuum.

Revision: Free Fall Equations

• $v^2 = u^2 + 2gs$

• $v = u + gt$

• $s=ut+\frac12gt^2$

Revision: Components of Vectors

• A vector is equal to the sum of its components, $v = v<em>x + v</em>y$

• Resolving a vector into its components.

• All vectors can be resolved into 2 perpendicular components:

  1. Horizontal component of force, $F_x = F cos \theta$

  2. Vertical component of force, $F_y = F sin \theta$

Projectile Motion Defined

• Projectile: projection (launching, throwing or firing) of an object at some angle into air and near the earth surface.

• Assumptions:

  • constant $g = 9.81 ms^{-2}$.

  • zero air resistance.

• If we neglect the effects of air resistance, then the path traced out by this projectile is a parabola.

Five Different Projectile Motion Scenarios

1. An object is projected at some angle above the horizontal and finishes at the same height.

2. An object is launched horizontally from some height and then falls to ground in parabolic path.

3. An object is projected at some angle above the horizontal, but the object finishes at a level below the initial launch height.

4. An object is projected at some angle above the horizontal, but the object finishes at a level above the initial launch height.

5. An object is projected from a height at some angle below the horizontal, i.e., it is thrown downwards.

Analyzing Projectile Motion

• We consider the horizontal component & vertical component separately.

• Direction of the velocity changes with time, v always at tangent to the parabolic path.

Magnitude of Velocity

• Magnitude of the velocity changes with time.

Horizontal Component of Velocity

• The horizontal component of velocity ($V_x$) remains constant throughout the motion.

Vertical Component of Velocity

• The vertical component of velocity ($v_y$) changes with time.

Projectile Motion at the Highest Point

• At the highest point of the projectile motion, vy = 0 $and$ v = vx only.

Velocity at Each Point

• $v$ at each point, v = Vx + Vy

Acceleration

• Acceleration at each point is constant ($a = -g$).

Horizontal Velocity Component

• Given a multiple image photograph of a projectile, can you deduce that horizontal velocity is constant?

• remains constant throughout the motion (if we neglect the effects of air resistance).

• Consider multiple exposures of projectile motion.

• The time interval between successive exposures is constant in each representation.

• The projectile travels equal distances horizontally between successive exposures.

• The horizontal component of the velocity of the projectile remains constant throughout the motion.

Vertical Velocity Component

• Given a multiple image photograph of a projectile, what can you deduce about the vertical velocity?

• changes during a projectile motion.

• Consider the vertical spacing between successive images.

• The vertical spacing between successive images decreases on the upward flight & increases on the downward flight.

• Thus, the vertical distance travelled between successive images in equal time interval varies.

• The vertical component of velocity changes during the motion.

Vertical Motion Comparison

• Consider two balls, released simultaneously.

• One ball falls vertically under the action of gravity.

• One ball is projected horizontally.

• Note that the vertical positions of the 2 balls are the same at all times.

• This means that the vertical velocities of both balls are the same at all times.

• The vertical component of a horizontally projected ball is identical to the motion of a ball, free-falling vertically downward.

• Both bodies have same initial vertical velocity component (zero) and experienced constant g downward (same acceleration in the vertical direction).

Projectile Motion Calculation

• Determination of:

  • Velocity of the projectile

  • Flight time

  • Range of the projectile

  • Maximum height attained by the projectile

Uni-level Angular Projection

• An object is projected with a speed, u at an angle above the horizontal, and finishes its flight at the same level.

  • Initial velocity components are: ux = u cos \theta$,$uy = u sin \theta

  • Examples of this type of motion are:

    • an athlete competing in the long jum

    • a golfer hitting a golf ball

    • a cricketer throwing a cricket ball to the wicketkeeper

Horizontal Projection

• A body is projected horizontally with speed, u from some height and finishes its flight at some level, h units below its initial height.

  • The initial horizontal velocity is the initial speed of the body, and the initial vertical velocity,

  • Examples of this type of motion are:

    • a ball rolling off a table top

    • a stone thrown from a cliff top

    • a car plunging off a cliff

  • $u_x = u = u cos 0° = u (1)$

  • $u_y = 0 = u sin 0° = u (0)$

Maximum Height Calculation

• Maximum height, H

  • Initial velocity:

  • Acceleration: $a = -g = -9.8 ms^{-2}$

  • At the maximum height, vertical component of velocity, $v_y = 0$

  • Using Vertical Calculations

    • $v^2 = u^2 + 2as$

    • $v<em>y^2 = u</em>y^2 + 2a<em>y s</em>y$

    • $0 = (u sin \theta)^2 + 2(-g)H$

    • Maximum height, $H = \frac{u^2 sin^2 \theta}{2g}$

    • where $sin^2 \theta = (sin \theta)(sin \theta)$

Time to Reach Maximum Height

• Using Vertical Calculations

  • $v = u + at$

  • $v<em>y = u</em>y + a_y t$

  • $0 = u sin \theta + (-g) t$

  • Time to reach the maximum height, $t = \frac{u sin \theta}{g}$

  • At the maximum height, $v_y = 0$

Instantaneous Height

• Using Vertical Calculations

  • $s = ut + \frac{1}{2} a t^2$

  • $s<em>y = u</em>y t + \frac{1}{2} (-g) t^2$

  • The instantaneous height, $h = u sin \theta t - \frac{1}{2} g t^2$

Time of Flight

• Time of flight, T (time taken by the projectile to return to the ground)

  • Using Vertical Calculations

    • $s = ut + \frac{1}{2} a t^2$

    • $s<em>y = u</em>y T + \frac{1}{2} -g T^2$

    • $0 = u sin \theta T - \frac{1}{2} g T^2$

    • Time of flight, $T = \frac{2 u sin \theta}{g} = 2t$

  • When the projectile returns to the ground, the vertical displacement, $s_y = 0$

Instantaneous Horizontal Distance

• The instantaneous horizontal distance travel, r at any time, t

  • Using Horizontal Calculation

    • $s = vt$

    • $s<em>x = u</em>x t$

    • $r = (u cos \theta) t$

Range Calculation

• The range, R (The maximum horizontal displacement)

  • Using Horizontal Calculation

    • $s = vt$

    • $s<em>x = u</em>x t$

    • $R = (u cos \theta)(\frac{2 u sin \theta}{g})$

    • $R = (u cos \theta) T$

    • $since T = \frac{2 u sin \theta}{g}$

    • $2 sin \theta cos \theta = sin 2 \theta$

    • Range, $R = \frac{u^2 sin 2\theta}{g}$

Maximum Range

• The range R is maximum when the angle of projection is 45°

  • To obtain the maximum range,

    • $sin 2 \theta = 1$

    • $R = \frac{u^2}{g}$

    • $2 \theta = 90°$

    • $\theta = 45°$

  • Range, $R = \frac{u^2 sin 2\theta}{g}$

Relationship Between Range and Launch Angle

• For a projectile of given speed, the range will be maximum if the launch angle is 45° above the horizontal.
  *Note: this best observed under a uni-level projection.

Conclusions About Projectile Range and Angle

1. For a projectile of given speed, the range will be maximum if the launch angle is 45° above the horizontal.

2. For two projectiles of given speed, the ranges will be the same if the two launch angles are symmetrical about 45°

3. For two projectiles of given speed, the ranges will be the same if the two launch angles add up to 90°

Example 1: Projectile Motion Calculation

• A projectile leaves the ground at an angle of 30° to the horizontal with an initial velocity of 20 ms-1.

• Assuming that the acceleration due to gravity is 9.81 ms-2, find the:

  • a) maximum height reached

  • b) time of flight

  • c) range of the object

Solution 1(a): Maximum Height

• Using $v^2 = u^2 + 2as$

  • $v<em>y^2 = u</em>y^2 + 2gH$

  • $0 = (u sin \theta)^2 + 2(-g)H$

  • At the maximum height, $v_y = 0$

  • $0 = (20 sin 30)^2 + 2(-9.81)H$

  • $0 = (10)^2 + (-19.62)H$

  • Maximum height, $H = \frac{u^2 sin^2 \theta}{2g} = \frac{(20)^2 (sin 30)^2}{2(9.81)} = 5.10 m$

  • Using H = 5.10 m

Solution 1(b): Time of Flight

• Using $s = ut + \frac{1}{2} a t^2$

  • $s<em>y = u</em>y T + \frac{1}{2} g T^2$

  • When the projectile returns to the ground, the vertical displacement, $s_y = 0$

  • $0 = (u sin \theta)T + \frac{1}{2} -g T^2$

  • $0 = 20 sin 30 T + \frac{1}{2} (-9.81)T^2$

  • $0 = 10 T + -4.91 T^2$

  • $4.91 T^2 – 10 T = 0$

  • $T (4.91 T – 10) = 0$

  • so T = 0 or 4.91 T – 10 = 0

  • ∴ T = 2.04 s

  • Time of flight, $T = \frac{2 u sin \theta}{g} = \frac{2 (20) sin 30}{9.81} = 2.04 s$

  • Using T = 2.04 s

Solution 1(c): Range of the Object

• $s_x = (u cos \theta)T$

  • $s<em>x = u</em>x t$

  • $s_x = (20 cos 30) (2.04) = 35.33 m$

  • Range, $R = \frac{u^2 sin 2\theta}{g}$

    • $R = \frac{(20)^2 sin (2x30)}{9.81} = 35.31 m$

Example 2: Motorcycle Stunt

• A motorcycle stunt-rider moving horizontally takes off from a point 5.0 m above the ground with a speed of 30 ms-1. How far away does the motorcycle land?

Solution 2

• Consider vertical component of motion:

  • $u_y = 0$

  • $a = + g$

  • $s_y = 5 m$

  • Using $s = ut + \frac{1}{2} a t^2$

    • $s<em>y = u</em>y t + \frac{1}{2} g t^2$

    • $5 = (0)t + \frac{1}{2} 9.81 t^2$

    • $t = 1.01 s$

  • $s<em>x = u</em>x t$

    • $s_x = (30)(1.01) = 30.29 m$

IMPORTANT FORMULAS!!!!!

1. Equations of Linear Motion

   • For an object moving with uniform velocity: $s = vt$

   • For an object moving with uniform acceleration:
     $s = ut + \frac{1}{2} a t^2$
     $v = u + at$
     $v^2 = u^2 + 2as$

2. Free Fall Equations

   • $v^2 = u^2 + 2gs$

   • $v = u + gt$

   • $s = ut + \frac{1}{2} g t^2$

3. Components of Vectors

   • Total vector: $v = v<em>x + v</em>y$

   • Horizontal component: $F_x = F \cos \theta$

   • Vertical component: $F_y = F \sin \theta$

4. Maximum Height Calculation for Projectile Motion

   • $H = \frac{u^2 \sin^2 \theta}{2g}$

5. Time to Reach Maximum Height

   • $t = \frac{u \sin \theta}{g}$

6. Time of Flight

   • $T = \frac{2 u \sin \theta}{g}$

7. Range Calculation

   • $R = \frac{u^2 \sin 2\theta}{g}$

8. Maximum Range

   • Occurs at $\theta = 45°$, with $R = \frac{u^2}{g}$

9. Instantaneous Horizontal Distance

   • $r = (u \cos \theta) t$