Module 3
Planning Guide

Lesson 1.10
. Try This
. Concepts
. Definitions
. Equations
. Examples
. Review
. Answers
. @home

Lesson 1.9
Lesson 1.10
Lesson 1.11
Lesson 1.12
Lab 1.9
Lab 1.10
Lab 1.11
Lab 1.12
Project 3


LESSON 1.10 - PROJECTILES

Objectives
On completion of this lesson, you should be able to:
Describe: The factors that affect the path of a projectile after it is launched. Describe the variations in kinetic energy and gravitational potential energy of a vertically launched projectile as it travels upwards and downwards.
Explain: Why a projectile that is launched vertically upwards will have nearly the same speed that it was launched with (but in the opposite direction) when it returns to the ground.
Calculate: The velocity and/or position of a projectile after a certain time. Calculate the distance traveled by a projectile in a certain time.


ACTIVITIES

  1. Monkey and Nerf Gun: Fire a projectile at an object that is released to fall as the projectile is launched.
  2. Determine the two angles at which a projectile can be launched at to achieve the same range.

Vector: A line, arrow or set of coordinates that represents a quantity and its direction.

Resultant: The combined effect of two or more vectors.

Projectile: An object that is given an initial velocity and then moves only under the influence of gravity.

Velocity Component: A portion of the velocity of an object that acts in a given direction.

Trajectory: The path followed by a projectile.

Projectiles
A projectile is an object that is given an initial velocity and then moves only under the influence of gravity. The height from which the projectile is launched, the velocity and the angle at which it is launched will determine the distance that it covers before striking the ground.

Horizontal and Vertical Components of Velocity
A horizontally launched projectile will start to drop as soon as it leaves the launching device. The time that it has to reach the ground is determined by its height. While it is moving, its velocity can be regarded as having two components: a vertical component and a horizontal component. In the absence of friction, the horizontal component remains constant at the value it had at launch. The vertical component of its velocity increases as the projectile accelerates towards the ground.

Acceleration due to gravity
If a projectile is launched upwards or at an angle to the horizontal, the vertical component of its velocity will come under the influence of gravity as soon as it leaves the launcher. Its vertical velocity away from the ground will be decreased by 9.81 meters per second per second (m/s2). If it travels towards the ground, its vertical velocity downwards will be increased by 9.81 m/s2.

Range
The distance that a projectile can cover before reaching the ground depends on the horizontal component of its velocity and the time that it has before reaching the ground. The time that it has before reaching the ground depends on its initial height and the vertical component of its velocity. The vertical component depends on the angle at which it is launched and its velocity.



v = at

Velocity = acceleration x time
Where: v = velocity
a = acceleration (M/s2)
t = time (s)

d = vt Distance = velocity x time

Where: d = distance (m)
v = velocity* (m/s)
t = time (s)
(*If v changes with time, use average velocity)

d = ½at2 Distance = ½ x acceleration x time2 (for accelerating object)

Where: d = distance (m)
a = acceleration (m/s2)
t = time (s)

ac = v 2/r Centripetal acceleration = (tangential velocity)2 ÷ radius of curve

Where: ac = centripetal acceleration (m/s2)
v = tangential velocity (m/s)
r = radius of curve (m)

For a projectile launched vertically from the ground at an initial
velocity of v0 :

v = v0 - gt

Velocity = initial upward velocity – (acceleration due to gravity x time)

Where: v = velocity (m/s)
v0 = initial velocity (m/s)
g = acceleration due to gravity (9.81 m/s2)
t = time (s)

tmax = v0 /g Time to reach maximum height = initial velocity ÷ acceleration due to gravity

Where: v0 = initial velocity (m/s)
g = acceleration due to gravity (9.81 m/s2)
tmax = time to reach maximum height (s)

dmax = ½g tmax 2 Maximum height = ½ x acceleration due to gravity x time2

Where: dmax = maximum height (m)
g = acceleration due to gravity (9.81 m/s2)
tmax = time to reach maximum height (s)

dmax = [(v0 + vt)/2] tmax

Maximum height = average velocity x time

Where: dmax = maximum height (m)
v0 = initial velocity (m/s)
vt = velocity at max height = 0 (m/s)
tmax = time to reach maximum height (s)

Example 1.10.1 Vertically launched projectile
A steel ball is dropped from a point that is 20 meters above ground level.

a) How fast will it be dropping 1 second after being dropped?
b) How far will it have dropped 1 second after being dropped?

Solution

  1. Velocity = acceleration x time. The acceleration due to gravity is 9.81 m/s2
  2. v = 9.81 m/s2 x 1 s = 9.81 m/s
  3. Distance = average velocity x time
  • The velocity at time = 0 is 0 m/s. The velocity at time = 1 s is 9.81 m/s

    The average velocity = ½ x 9.81 = 4.905 m/s

    The distance covered is thus 4.905 m/s x 1 s = 4.905 m

    •


    Example 1.10.2 Vertically launched projectile
    A baseball is thrown vertically and lands 2 seconds later.

    1. What height does it reach?
    2. At what velocity does it strike the ground.

    (Assume that it is launched from ground level and neglect air resistance)

    Solution

    a) The time taken to reach the maximum height is the same as the time taken to reach the ground from the maximum height. tmax = 1 second.

  • Use the equation: d = ½gt2 : Maximum height = ½ x acceleration due to gravity x time2

    Maximum height = 0.5 x 9.81 m/s2 x 12 = 4.905 m.

    •

    b) tmax = v0 /g therefore v0 = tmax g

    The velocity at which the ball was launched was:

    v0 = 1 s x 9.81 m/s2 = 9.81 m/s

    This is the same velocity at which it strikes the ground – but in the opposite direction.


    Example 1.10.3 Projectile launched at an angle
    A baseball is thrown from a height of 1.6 meters with a velocity of 120 kilometers per hour at an angle of 20º upwards of the horizontal.

    How far will it travel before reaching the ground? (Neglect air resistance)

    Solution

    The initial velocity of the ball is 120,000 /(60 x 60) = 33.3 m/s

    The horizontal component of this velocity is 33.3 x cos 20º = 33.3 x 0.94 = 31.3 m/s

    The vertical component of the initial velocity = 33.3 x Sin 20º = 33.3 x 0.342 = 11.4 m/s

    To calculate the time it has before reaching the ground, we calculate the time taken to reach its maximum height and add to this the time to drop from the maximum height – remembering that it is launched from a height of 1.6 meters.

    Time to reach maximum height (above 1.6m): ) tmax = v0 /g = 11.4 / 9.81
    = 1.16 seconds.

    The maximum height (above 1.6m) : dmax = ½g tmax 2
    = 0.5 x 9.81 x (1.16)2 = 6.6 meters.

    The maximum height = 1.6 m + 6.6 m = 8.2 meters.

    The time to drop to the ground from 8.2 m :

    Rearrange the equation d = ½gt2 (g is the acceleration in place of a)

    . . . .____ . . .___________
    t =     Ö 2d/g = Ö 2 x 8.2 / 9.81 = 3.13 seconds.

    The total time that the ball is in the air is 3.13 + 1.16 = 6.4 seconds.

    With a horizontal velocity of 31.3 m/s it can reach a distance
    of 31.3 x 6.4 = 200.2 meters in this time.

    Back to Top of Page

    Review Questions

    1. What is a trajectory?
    2. Does a projectile that is launched horizontally from a height of 10 meters take longer to reach the ground if it is launched with a greater speed? Explain.
    3. Why does an arrow that is launched vertically hit the ground at almost the same speed as it left the bow?
    4. Is it true that an arrow launched vertically upwards starts to accelerate towards the ground as soon as it leaves the bow?
    5. An arrow with a mass of 200 grams leaves the bow with an upward velocity of 4 m/s:
      Relative to the point that it was launched from, how much gravitational potential energy does it have when it reaches it’s maximum height?
      What is the maximum height above the launching point?
    6. An arrow is fired horizontally with a velocity of 4 m/s from a point 5 meters above the ground:
      How long will it take to reach the ground?
      How far will it travel relative to the ground before hitting the ground?
    7. An arrow is fired from a point 2 meters above the ground with a velocity of 5 m/s at an angle of 30º to the horizontal:
      What are the vertical and horizontal components of it’s velocity?
      How long will it take to reach the ground?
      How far will it travel relative to the ground before hitting the ground?

    Back to Top of Page

    CHOICE OF ACTIVITIES

    1. Monkey and Nerf Gun: Fire a projectile at an object that is released to fall as the projectile is launched.
    2. Determine the two angles at which a projectile can be launched at to achieve the same range.

    ACTIVITY 1.10.1 - Monkey and Nerf gun
    Purpose:     To show that a projectile starts to fall as soon as it leaves the launcher.

    Equipment:
    A Nerf gun or similar projectile launcher that is relatively safe to use in a classroom

    A toy monkey – or substitute object that can be dropped without causing damage or breaking.

    Procedure:

    1. One volunteer holds the monkey while standing near the top of a step-ladder or on a sturdy table.
    2. A second volunteer aims the Nerf gun directly at the monkey.
    3. Arrange an appropriate signal so that the Nerf gun is fired as the monkey is dropped.
    4. If the gun is aimed correctly, the projectile should hit the monkey every time – irrespective of the distance from the monkey.

    Question

    1. Why does a correctly aimed projectile hit the falling object?

    ACTIVITY 1.10.2 - Water Hose Angles
    Purpose:     To show that there are usually two angles at which a projectile can be launched in order to reach the same point.

    Equipment:
    Watering hose with a nozzle.

    Procedure:

    1. Place a bucket or container about 4 meters away.
    2. See if a stream of water from the hose can be aimed into the container from two different angles.

    .

    HANDS-ON HOMEWORK
    Select one or more of the     recommended activities for Lesson 1.11, collect the items needed and test the procedure before demonstrating the activity during the next theory lesson.

    Lesson 1.10 PROJECTILES

    1. The path that a projectile follows.
    2. Yes. The time taken to reach its maximum height depends on its upward velocity when it is launched.
    3. Apart from some loss of energy due to friction, the kinetic energy of the arrow is converted to gravitational potential energy as it travels upwards. This is converted back to kinetic energy on the way down.
    4. Yes, it experiences negative acceleration due to gravity.
    5. Its kinetic energy at launch = ½mv2 = 0.5 x 0.2kg x 42 = 1.6 J
      This is all converted to gravitational potential energy = 1.6 J.
      If gravitational potential energy = mgh, h = 1.6 / (9.81 x 0.2) = 0.81 m
    6. The time to drop to the ground from 5 m :
      Rearrange the equation       d = ½gt2 (g is the acceleration in place of a)
      . . . .____ . . .___________
      t =       Ö 2d/g = Ö 2 x 5 / 9.81 = 1.01 seconds.

      The distance covered = 4 x 1.01 = 4.04m.
    7. The vertical component of its velocity = 5m/s x sin 30º = 2.5 m/s
      The horizontal component of its velocity = 5 x cos 30º = 4.33 m/s
      Time to reach maximum height (above 2m): )       tmax = v0 /g = 2.5 / 9.81 = 0.255 seconds.

    The maximum height (above 2m) : dmax = ½g tmax 2 = 0.5 x 9.81 x (0.255)2 = 0.32 meters.

    The maximum height = 2 m + 0.32 m = 2.32 meters.

    The time to drop to the ground from 2.32 m :

    Rearrange the equation d = ½gt2 (g is the acceleration in place of a)

    . . . .____ . . .___________
    t =     Ö 2d/g = Ö 2 x 2.32 / 9.81 = 0.688 seconds.

    The total time that thearrow is in the air is 0.255s + 0.688 = 0.943 seconds.

    With a horizontal velocity of 4.33 m/s it can reach a distance of 4.33 x 0.943 = 4.08 meters in this time.