Derivation of the equations of motion of a projectile

1. PROJECTILES
MUJUNGU HERBERT
(Mathematics Lecturer)
National Teachers College Kabale
May 25, 2020

2. Content
1 Definition of a projectile
Examples
2 Equations of motion for a projectile
3 Other general equations
Time at maximum height
Maximum Height Hmax
Time of flight T
Range R
Maximum range R
Equation of a trajectory
4 Summary
MUJUNGU HERBERT (Mathematics Lecturer) (National Teachers College Kabale)PROJECTILES May 25, 2020 2 / 13

3. Definition of a projectile
Any Object that once thrown by some force continues in motion by its
own inertia.
Examples
Throwing/kicking a ball,
Throwing a stone,
A shell fired from a gun
etc
MUJUNGU HERBERT (Mathematics Lecturer) (National Teachers College Kabale)PROJECTILES May 25, 2020 3 / 13

4. A particle projected at an angle θ to the horizontal
u
θ
u cos θ
u sin θ
g
X-axis
Y-axis
vx
vy
O
We shall investigate both the horizontal and vertical motions;
For horizontal motion, vx = ucosθ is constant since there are no
horizontal forces subject to the particle,
The vertical component of velocity, vy, is subject to acceleration due to
gravity, g.
MUJUNGU HERBERT (Mathematics Lecturer) (National Teachers College Kabale)PROJECTILES May 25, 2020 4 / 13

5. Equations of motion for a projectile
The components of acceleration parallel to the axes at any instant are;
ax = 0 , ay = −g (1)
Integrating Equations 1 with respect to time, we have;
vx = constant , vy = −gt + constant
At t = 0, vx and vy are ucosθ and usinθ respectively, hence;
vx = ucosθ , vy = −gt + usinθ (2)
Integrating again with respect to time;
x = ucosθ · t , y = −1
2gt2 + usinθ · t (3)
Equations 2 and 3 give the velocity and coordinates respectively of a
projectile at any instant. Also; v2
y = u2
sin2
θ · −2gy, can be used.
MUJUNGU HERBERT (Mathematics Lecturer) (National Teachers College Kabale)PROJECTILES May 25, 2020 5 / 13

6. Time at maximum height
Taking the fact that the particle is
momentarily at rest at maximum
height and by using Equation 2;
vy = −gt + usinθ
But vy = 0
0 = −gt + usinθ
t =
u sin θ
g
(4)
Maximum Height Hmax
Using the time at maximum height
and 3, i.e,
y = −
1
2
gt2
+ u sin θt
Hmax = −
1
2
g
u sin θ
g
2
+ u sin θ
u sin θ
g
Hmax =
u2
sin2
θ
2g
(5)
MUJUNGU HERBERT (Mathematics Lecturer) (National Teachers College Kabale)PROJECTILES May 25, 2020 6 / 13

7. Time of flight T
For a complete flight, y = 0 and
t = T;
By using Eqn 3;
0 = −
1
2
gT2
+ u sin θT
T(u sin θ −
1
2
gT) = 0
T = 0 orT =
2u sin θ
g
Ignoring T = 0 then
T =
2u sin θ
g
(6)
Range R
R = T × u cos θ
R =
2u sin θ
g
u cos θ
R =
2u2 sin θ cos θ
g
R =
u2 sin 2θ
g
(7)
This is maximum when
sin 2θ = 1, i.e, is maximum;
Rmax =
u2
g
when θ = 45o (8)
MUJUNGU HERBERT (Mathematics Lecturer) (National Teachers College Kabale)PROJECTILES May 25, 2020 7 / 13

8. Equation of a trajectory
From equation 3; x = ucosθ · t , y = −1
2gt2 + usinθt,
(a, b)
θ1
θ2
u
u
x1
x2
y y
(x1, y) (x1, y)
u
θ
1 Show that the
Equation of the trajectory is
y = x tan θ −
gx2
2u2
sec2
θ.
2 If the particle
passes through the point with
coordinates (a, b), show that
ga2
tan2
θ − 2u2
a tan θ + 2u2
b + ga2
= 0
MUJUNGU HERBERT (Mathematics Lecturer) (National Teachers College Kabale)PROJECTILES May 25, 2020 8 / 13

9. Example 1
A particle is projected from a point on level ground such that its initial
velocity is 56 ms−1 at an angle of 30 o above the horizontal. Taking g as
9.8 ms−2, find the time taken for the particle to reach its maximum height.
Solution
MUJUNGU HERBERT (Mathematics Lecturer) (National Teachers College Kabale)PROJECTILES May 25, 2020 9 / 13

10. Example 1
A particle is projected from a point on level ground such that its initial
velocity is 56 ms−1 at an angle of 30 o above the horizontal. Taking g as
9.8 ms−2, find the time taken for the particle to reach its maximum height.
Solution
Taking vertical motion and using the fact that the particle is momentarily
at rest vertically at the maximum height;
MUJUNGU HERBERT (Mathematics Lecturer) (National Teachers College Kabale)PROJECTILES May 25, 2020 9 / 13

11. Example 1
A particle is projected from a point on level ground such that its initial
velocity is 56 ms−1 at an angle of 30 o above the horizontal. Taking g as
9.8 ms−2, find the time taken for the particle to reach its maximum height.
Solution
Taking vertical motion and using the fact that the particle is momentarily
at rest vertically at the maximum height;
Applying
˙y = 0, u = 56 ms−1, θ = 30 o and
g = 9.8 ms−2 in equation 2
MUJUNGU HERBERT (Mathematics Lecturer) (National Teachers College Kabale)PROJECTILES May 25, 2020 9 / 13

12. Example 1
A particle is projected from a point on level ground such that its initial
velocity is 56 ms−1 at an angle of 30 o above the horizontal. Taking g as
9.8 ms−2, find the time taken for the particle to reach its maximum height.
Solution
Taking vertical motion and using the fact that the particle is momentarily
at rest vertically at the maximum height;
Applying
˙y = 0, u = 56 ms−1, θ = 30 o and
g = 9.8 ms−2 in equation 2 , i.e,
0 = −gt + u sin θ
0 = −9.8t + 56sin30o
t =
56sin30o
9.8
, ⇒ t = 2.86 s
MUJUNGU HERBERT (Mathematics Lecturer) (National Teachers College Kabale)PROJECTILES May 25, 2020 9 / 13

13. Example 2
A bullet fired from a gun has a maximum horizontal range of 2000 m. Find
the muzzle velocity of the gun (i.e. the speed with which the bullet leaves
the gun).
Solution
MUJUNGU HERBERT (Mathematics Lecturer) (National Teachers College Kabale)PROJECTILES May 25, 2020 10 / 13

14. Example 2
A bullet fired from a gun has a maximum horizontal range of 2000 m. Find
the muzzle velocity of the gun (i.e. the speed with which the bullet leaves
the gun).
Solution
MUJUNGU HERBERT (Mathematics Lecturer) (National Teachers College Kabale)PROJECTILES May 25, 2020 10 / 13

15. Example 2
A bullet fired from a gun has a maximum horizontal range of 2000 m. Find
the muzzle velocity of the gun (i.e. the speed with which the bullet leaves
the gun).
Solution
So we have θ = 45o,
Rmax = 2000 m, u =? and
g = 9.8 ms−2
MUJUNGU HERBERT (Mathematics Lecturer) (National Teachers College Kabale)PROJECTILES May 25, 2020 10 / 13

16. Example 2
A bullet fired from a gun has a maximum horizontal range of 2000 m. Find
the muzzle velocity of the gun (i.e. the speed with which the bullet leaves
the gun).
Solution
So we have θ = 45o,
Rmax = 2000 m, u =? and
g = 9.8 ms−2 By using in Eqn 8;
Rmax =
u2
g
2000 =
u2
9.8
u = (2000 ∗ 9.8), ⇒ u = 140 ms−1
MUJUNGU HERBERT (Mathematics Lecturer) (National Teachers College Kabale)PROJECTILES May 25, 2020 10 / 13

17. Trial Questions
1 A particle is projected from a point on level ground such that its
initial velocity is 28 ms−1 at an angle of 45 o above the horizontal.
Taking g as 9.8 ms−2, find the time taken for the particle to reach its
maximum height.
MUJUNGU HERBERT (Mathematics Lecturer) (National Teachers College Kabale)PROJECTILES May 25, 2020 11 / 13

18. Trial Questions
1 A particle is projected from a point on level ground such that its
initial velocity is 28 ms−1 at an angle of 45 o above the horizontal.
Taking g as 9.8 ms−2, find the time taken for the particle to reach its
maximum height.
2 A particle is projected from a point on a horizontal plane and has an
initial velocity u at an angle of θ above the plane. Show;
by using the equation; v2
y = u2
sin2
θ − 2gy, that the greatest height h
reached by the particle above the plane is given by h = u2 sin2 θ
2g .
MUJUNGU HERBERT (Mathematics Lecturer) (National Teachers College Kabale)PROJECTILES May 25, 2020 11 / 13

19. Trial Questions
1 A particle is projected from a point on level ground such that its
initial velocity is 28 ms−1 at an angle of 45 o above the horizontal.
Taking g as 9.8 ms−2, find the time taken for the particle to reach its
maximum height.
2 A particle is projected from a point on a horizontal plane and has an
initial velocity u at an angle of θ above the plane. Show;
by using the equation; v2
y = u2
sin2
θ − 2gy, that the greatest height h
reached by the particle above the plane is given by h = u2 sin2 θ
2g .
3 A particle is projected from a point on a horizontal plane with an
initial velocity of 140 ms−1 at an angle of elevation of 30o. Find the
greatest height reached by the particle above the plane.
MUJUNGU HERBERT (Mathematics Lecturer) (National Teachers College Kabale)PROJECTILES May 25, 2020 11 / 13

20. Trial Questions
1 A particle is projected from a point on level ground such that its
initial velocity is 28 ms−1 at an angle of 45 o above the horizontal.
Taking g as 9.8 ms−2, find the time taken for the particle to reach its
maximum height.
2 A particle is projected from a point on a horizontal plane and has an
initial velocity u at an angle of θ above the plane. Show;
by using the equation; v2
y = u2
sin2
θ − 2gy, that the greatest height h
reached by the particle above the plane is given by h = u2 sin2 θ
2g .
3 A particle is projected from a point on a horizontal plane with an
initial velocity of 140 ms−1 at an angle of elevation of 30o. Find the
greatest height reached by the particle above the plane.
For more questions; https://bit.ly/Projectile_questions
and for notes https://bit.ly/Projectile_notes .
All are reading materials are saved on https://bit.ly/padlet-DESII
MUJUNGU HERBERT (Mathematics Lecturer) (National Teachers College Kabale)PROJECTILES May 25, 2020 11 / 13

21. Summary
A Projectile is any object that when thrown continues, continues in
motion under its own inertia.
Equations of motion
1 Horizontal Motion;
x = u cos θ · t
2 Vertical motion;
vy = u sin θ − gt
y = −
1
2
gt2
+ u sin θ · t and
v2
y = u2
sin2
θ − 2gy
1 Maximum Height;
Hmax =
u2 sin2
θ
2g
,
2 Time at Hmax; t = u sin θ
g ,
3 Time of Flight; T =
2u sin θ
g
,
4 Range;R =
u2 sin 2θ
g
,
Rmax =
u2
g
Equation of a Trajectory; y = x tan θ −
gx2
2u2
sec2
θ.
MUJUNGU HERBERT (Mathematics Lecturer) (National Teachers College Kabale)PROJECTILES May 25, 2020 12 / 13

22. Thank You for Your Attention
For more information Contact me through; herbertmujungu@gmail.com
+256779547251
+256701310635
+256793854372
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2019/2020
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NTC-KABALE, MTC2, 2019/20
MUJUNGU HERBERT (Mathematics Lecturer) (National Teachers College Kabale)PROJECTILES May 25, 2020 13 / 13