A projectile is an object moving under the influence of gravity with a parabolic path. Its motion can be analyzed by separating the horizontal and vertical components. In the horizontal direction, the velocity is constant, while in the vertical direction there is a constant acceleration due to gravity. Projectile motion is used to model many real-world scenarios like thrown objects, diving, and artillery fire. Solving projectile motion problems involves separating the horizontal and vertical motions and using kinematic equations with the initial velocities and gravitational acceleration.

1. LECTURE 5
Projectile Motion
IB Physics Power Points
Topic 9.1
Motion in Fields
www.pedagogics.ca

2. Projectile Motion Is Everywhere

3. A projectile is an object moving in two dimensions under
the influence of only Earth's gravity; its path is a
parabola.

4. Projectile motion can be
understood by analyzing the
horizontal and vertical
motions separately.

5. The velocity in the x-direction is
constant;
in the y-direction the object has
a constant acceleration g.
This photograph shows two balls that
start to fall at the same time. The one
on the right has an initial horizontal
velocity.
It can be seen that vertical positions
of the two balls are identical at
identical times, while the horizontal
position of the yellow ball increases
linearly.

6. Key Concept:
A projectile is a free falling object that moves
sideways.
Vertically, a projectile is no different than a
dropped object or an object that has been
thrown straight upwards or downwards.
g = - 9.81 ms-2

7. Key Concept:
A projectile is a free falling object that moves sideways.
Horizontally, a projectile is no different than any other
object moving with a constant velocity. Gravity has NO
effect on the horizontal motion. a = 0 ms-2

8. A horizontally launched projectile will have an initial
vertical velocity of zero. It is a dropped object moving
sideways.
Compare the flight time for each of these three paths:

9. A diver running 1.8 ms-1 dives out horizontally from the
edge of a vertical cliff and 3.0 s later reaches the water
below. How high was the cliff, and how far from its base did
the diver hit the water?
Δsy = uyΔt + ½gΔt2
= 0 + ½(-9.81)(3.0)2
= - 44 m
Δsx = uxΔt
= (1.8)(3.0)
= 5.4 m

10. If an object is launched at an initial angle of θ with the
horizontal, the analysis is similar except that the initial
velocity now has a vertical component.

11. A projectile is fired with an initial speed of 65.2 ms-1 at an
angle of 34.5°above the horizontal on a long flat firing range.
Determine (a) the maximum height reached by the projectile,
(b) the total time in the air, (c) the total horizontal distance
covered (that is, the range), and (d) the velocity of the
projectile 1.50 s after firing.

12. Solving Problems Involving Projectile Motion
Projectile motion is motion with constant acceleration
in two dimensions.
ax= 0 ay = g = -9.81 ms-2
Horizontal motion Vertical motion
vx = ux vy = uy+gΔt
sx = uxΔt Δsy = uyΔt +½gΔt2
v2
y= u2
y +2g(Δsy )

13. 1. Read the problem carefully. Draw a diagram.
2. Choose an origin and a coordinate system.
3. Decide on the time interval; this is the same in both
directions, and includes only the time the object is
moving with constant acceleration g.
4. Examine the x and y motions separately.
5. List known and unknown quantities.
Remember that ux never changes, and that vy =
0 at the highest point.