College Preparatory Program • Saudi Aramco
Projectile Motion
Projectile Motion
If we kick or throw a ball it moves in two dimensions.
t1
t2
The motion can be represented graphically in the xy –plane. Since both x and y axes are
being used for position, the time information has to be given along the path.
A small object (a point without dimensions) was thrown from the origin (0, 0) of this
coordinate system with the initial velocity, 𝒗𝟎 , at an angle α from the positive direction
of x axis.
The projectile moves along the trajectory shown above. This trajectory is parabolic.
The motion of this object is studied by investigating the motion in the horizontal
direction and the vertical direction separately.
If we neglect air resistance the ball moves in the horizontal direction at constant velocity
𝑥
vx= v0 cosa =
𝑡
The object accelerates in the y-direction with constant acceleration g = -9.8 m/s2.
The initial velocity in the y-direction is
v0y= v0 sina
The motion along the x axis is completely independent from the motion along the y axis.
The distance traveled along the x axis as a function of time
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College Preparatory Program • Saudi Aramco
Projectile Motion
x(t) = v0cosα t
Distance travelled along the y axis is
1
𝑦 𝑡 = 𝑣0𝑦 𝑡 − 𝑔𝑡 2
2
Since the minus sign has been inserted in the equation g takes the positive value 9.8 m/s2
and velocity along the y axis is given by the equation
vy (t) = v0y – g t
There are two main questions asked for such a case of projectile motion:
1) What is a maximum vertical displacement for a given angle ,
2) What is a maximum horizontal displacement for a given angle .
The maximum vertical displacement is H. It can be calculated by determining the time it
takes to reach the maximum vertical displacement (i.e. when the velocity in the vertical
direction is zero).
0 = v0y –g t
So
𝑡=
𝑣0𝑦 𝑣0 𝑠𝑖𝑛𝛼
=
𝑔
𝑔
Substituting this time into the vertical displacement, we get the maximum horizontal
displacement of the projectile
𝑣0 2 𝑠𝑖𝑛2 𝛼
𝐻=
2𝑔
To find the maximum horizontal displacement for a given angle we must calculate the
total time projectile will be above the level of the x axis. From symmetry this is just
𝑡=
2𝑣0 𝑠𝑖𝑛𝛼
𝑔
Substituting this time in the equation for the largest horizontal displacement, R:
2𝑣0 2 𝑠𝑖𝑛𝛼𝑐𝑜𝑠𝛼 𝑣0 2 𝑠𝑖𝑛2𝛼
𝑅=
=
𝑔
𝑔
The next question we can ask is: for which angle , with a given initial velocity v0, will
a) the maximum vertical displacement, H, have its maximum value,
b) the maximum horizontal displacement , R, have its maximum value.
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College Preparatory Program • Saudi Aramco
Projectile Motion
We can use calculus for solving these questions. To find the maximum value of H we
find the derivative of this function with respect to , set it to zero and solve the equation.
𝑑𝐻 𝑣0 2 2𝑠𝑖𝑛𝛼𝑐𝑜𝑠𝛼
=
=0
𝑑𝛼
2𝑔
which is the condition for the existence of a maximum or minimum of a function. Solving
it with respect to α gives
α = 00 and α = 900
The first solution, α = 00 leads to vertical displacement equal to zero – the minimum
vertical displacement.
The maximum vertical displacement is obtained for the angle α = 900, i.e. sinα = 1
𝑣0 2
𝐻=
2𝑔
The angle at which the maximum horizontal displacement is obtained can be found
similarly.
𝑑𝑅
2𝑣0 2 𝑐𝑜𝑠2𝛼
=
= 0 𝑓𝑜𝑟 𝑚𝑎𝑥/𝑚𝑖𝑛
𝑑𝑡
𝑔
2α = 900,
α = 450.
maximum horizontal displacement is obtained when firing at an angle of 450.
Now we show the shape of the trajectory in projectile motion is parabolic. For this
purpose we will express vertical displacement y as a function of horizontal displacement
x, that is we will calculate a function of the type,
y = f(x)
𝑥 = 𝑣0 𝑐𝑜𝑠𝛼 𝑡
1
𝑦 = 𝑣0 𝑠𝑖𝑛𝛼 𝑡 − 𝑔𝑡 2
2
Eliminate t:
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College Preparatory Program • Saudi Aramco
𝑦=
Projectile Motion
𝑠𝑖𝑛𝛼
𝑔
𝑥−
𝑥2
2
2
𝑐𝑜𝑠𝛼
2𝑣0 𝑐𝑜𝑠 𝛼
and we write it in the form
y= b x + c x2
with
𝑏=
𝑠𝑖𝑛𝛼
𝑐𝑜𝑠𝛼
and
𝑐=−
𝑔
2𝑣0 2 𝑐𝑜𝑠 2 𝛼
Therefore the trajectory of the projectile motion is a PARABOLA.
We can ask the question – for which value of horizontal displacement x, the vertical
displacement y achieves its maximum?
𝑑𝑦
= 𝑏 + 2𝑐𝑥 = 0
𝑑𝑥
𝑏
𝑣0 2 𝑠𝑖𝑛𝛼𝑐𝑜𝑠𝛼 𝑣0 2 𝑠𝑖𝑛2𝛼
𝑥=− =
=
2𝑐
𝑔
2𝑔
Notice, that it is exactly half of the largest horizontal displacement, showing that the
motion is symmetric. The maximum height is at the half of the maximum displacement,
for any angle α .
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