Physics Tutorial: Motion in Two Dimensions. Projectile Motion

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In this Physics tutorial, you will learn:

• What is a projectile?
• How to divide the study of a projectile?
• What kinds of motion are involved in a projectile?
• What are the equations used during the study of a projectile?
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3.12Motion in Two Dimensions. Projectile Motion

Introduction

If you throw a stone vertically upwards, it eventually will fall on the ground after reaching a certain height. In this case, the object makes a one-dimensional vertical motion. We have discussed this situation in the Physics tutorial "The Meaning of Acceleration. Constant and Non-Constant Acceleration. Gravitational Acceleration." We have stated that this kind of motion is decelerated when the object is moving upwards and accelerated when it is moving downwards (when it is falling down). It is obvious the acceleration produced in this case is caused by the gravity. Therefore, (as stated in the abovementioned tutorial), it is also known as the "gravitational acceleration" or the "acceleration of free fall."

However, very rarely occurs that an object thrown upwards falling again at the starting point. It is difficult to have such a precision even if you try a lot. In most cases, the object will fall at another place, so its trajectory is not purely vertical. This tutorial is dedicated to such situations i.e. it deals with objects thrown at a certain angle to the horizontal direction that is different from 90° (because if the throwing angle was 90° to the horizon, the motion would be only vertical).

What is a projectile?

Projectile motion is a kind of motion experienced by an object that is projected near the Earth's surface and moves along a curved path under the action of gravity only (if the effects of air resistance are assumed to be negligible). This curved path is a parabola. In the special case when the object is thrown directly upwards as discussed before, it may also be a straight line. The study of such motions is called ballistics, and such a trajectory is known as a ballistic trajectory. As stated before, the parabolic trajectory takes place when the object is thrown at an angle that is different from 90° to the horizontal direction. Look at the figure below:

The object is thrown at initial velocity v0 and at an angle θ to the horizontal direction. The direction of the initial velocity vector is according the tangent to the line at the given point. Therefore, it may not show the direction of motion but touches the trajectory only at the starting point. It is obvious the angle θ is different from 90°, otherwise, the object would move only vertically. It cannot maintain the initial linear direction as shown by the vector v0 as the gravity pulls it down. As a result, the object will follow a parabolic path (it was first demonstrated by Galileo Galilei) as shown in the figure above.

When the object is thrown at an angle θ to the horizontal direction, the only downwards force is the gravitational force, which tries to pull the object down. However, the object continues to move up for a while due to the effect of the initial throwing force. Therefore, it will not fall immediately on the ground but it will follow a parabolic path. This parabolic path is otherwise known as the "trajectory" of the object.

The Physics of Projectile

As stated in the Physics tutorial "Vectors and Scalars", when we have a situation that involves at least two directions, we split the vector quantities involved in components. In this case, the initial velocity v0 is a vector quantity. Therefore, it is written in components where v0x represents its horizontal component and v0y the vertical one. This method of splitting the initial velocity into components is also used for another major reason. In absence of air resistance, the only acceleration acting on a projectile is the gravitational acceleration g. We know that it is a vertical (downward) acceleration. Therefore, the horizontal part of the motion does not contain any acceleration (it is a uniform motion). Hence, it is necessary to study the horizontal and the vertical components of projectile separately because the horizontal component represents a uniform motion and the vertical component a uniformly decelerated motion when moving up and a uniformly accelerated motion when falling down.

If we denote the horizontal direction by x and the vertical one by y, we have for the initial velocity v0:

v0x = |v0| × cos θ
v0y = |v0| × sin θ

If the total time of flight from the throwing position (if it is on the ground) until it falls again on the ground is ttot, we obtain for the horizontal displacement 0x∆x:

∆x = v0x × ttot
= v0 × cos θ × ttot

Look at the figure:

On the other hand, since the vertical part of motion is a motion with constant acceleration g, we can use the four known kinematic equations to find any missing quantity required. Thus, we have

Equation 1

vy = v0y + g × t
vy = v0 × sin θ + g × t

Or

Equation 2

∆y = (vy + v0y) × t/2
∆y = (vy + v0 × sin θ) × t/2

Or

Equation 3

v2y - v20y = 2 × g × ∆y
v2y - (v0 × sin θ )2 = 2 × g × ∆y

Or

Equation 4

∆y = v0y × t + g × t2/2
∆y = v0 × sin θ × t + g × t2/2

Remarks!

∆y here represents the change in the vertical position. It is equal to ∆y = y-y0 where y0 is the initial vertical position and y is the vertical position at the instant t. We often refer to ∆y as the height h (look at the Physics tutorial "The Meaning of Acceleration. Constant and Non-Constant Acceleration. Gravitational Acceleration."

We substitute the components v0x and v0y of the initial velocity by v0 × cos θ and v0 × sin θ respectively but we cannot assign a known angle to the velocity v at any instant t because the angle changes at every instant. Therefore, we simply write vy for the vertical component of the velocity vector v (the horizontal component vx is known; it is vx = v0x = |v0 | × cos θ during the entire motion as the horizontal component of a projectile is a uniform motion)

When the object reaches its highest position (at ∆ymax or hmax), it stops rising up and prepares to fall down. In that instant it is moving only horizontally as its vertical component of the velocity vy is zero. Thus, if we denote the highest point of a projectile trajectory by A, we have vyA = 0 and vxA = |vA | = v0 × cos θ.

The sign of the gravitational acceleration is taken the same as it was at the beginning of motion. Thus, if the object was thrown from the ground, the sign of g is taken as negative during the entire motion, despite the fact that it accelerates in the second part of the trajectory, i.e. when falling down.

The equation of motion for the vertical part of a projectile is the fourth (the last) equation listed above. Thus, we have

∆y = v0 × sin θ × t + g × t2/2
y - y0 = v0 × sin θ × t + g × t2/2

Or

y = y0 + v0 × sin θ × t + g × t2/2

Therefore, the vertical position y as a function of time t is

y(t) = y0 + v0 × sin θ × t + g × t2/2

It allows us to determine the vertical position of an object at every instant t when the initial parameters (the initial vertical coordinate y0 and the initial velocity v0) are known.

The figure below includes all the abovementioned details:

If we are interested to find the direction of the velocity vector at any given instant, we can use the concept of the tangent of the angle formed by the velocity vector and the horizontal direction. Hence, if we denote this angle by α (remember, it is different from the initial angle θ), we can write:

tan α = vy/vx

A small portion of the trajectory is shown in the figure below to illustrate this point:

If the final vertical position of the object is the same as the initial one, there is a symmetry in the trajectory where the vertical line starting from the highest position can be considered as the line of symmetry of parabola. In this case, as discussed in the Physics tutorial "The Meaning of Acceleration. Constant and Non-Constant Acceleration. Gravitational Acceleration," the relationship between the rising time tup, the falling time tdown and the total time of flight ttotal is

tup = tdown = ttotal/2

The object is in two different instants at the same vertical position (height) during a projectile motion: one when it is rising up and the other when it is falling down. Unlike in the pure vertical motion, the object is in different horizontal positions in these two instants, so we have two distinct points M and N in the space in which the object is at the same height. Their coordinates are M(xM, yM) and N(xN, yN) respectively, where yM = yN. Look at the figure:

(The values obtained for the horizontal coordinates are rounded to 1 decimal place)

Let's consider at an example using the clues of the trajectory shown in the above figure.

Example 1

An object is moving according to a ballistic trajectory as shown in the figure.

Calculate:

1. The total time of flight
2. The magnitude of the initial velocity
3. The angle θ formed by the initial velocity vector and the horizontal direction
4. The coordinates of the object at t = 0.3 s
5. The instant(s) in which the object is at y = 3m

Solution 1

We can obtain some useful information from the figure. Thus, we see that the initial position of the object is at (0, 0). This means x0 = 0 and y0 = 0.

Also, it is obvious the object reaches its maximum position at yA = ymax = 4m. The corresponding horizontal coordinate is xA = 2m.

Yet, the maximum horizontal displacement is Δxmax = 4m as the object falls on the ground at that horizontal position.

Before starting with the solution, let's complete the figure with the data obtained above. (For simplicity we take the magnitude of gravitational acceleration equal to 10 m/s2.)

We use the info obtained for the point A to calculate the total time of flight ttot. We have:

v2yA - v20y = 2 × g × ∆ymax

Substituting the values, we obtain

02 - v20y = 2 × (-10) × 4
-v20y = -80
v0y = √80m/s ≈ 8.94 m/s

Now, we can calculate the rising time tup. We have:

∆ymax = (vyA + voy)tup/2

Substituting the values (and giving that vyA = 0), we obtain

4 = (0 + √80 × tup/2
8 = √80 × tup
tup = 8/80= 8 × √8/8 × √10= √8/10 = 0.89 s

Therefore, by symmetry it is obvious the total time of flight is

ttot = 2 × tup = 2 × 0.89 s = 1.78 s

We need to know the horizontal component of the initial velocity v0x in order to calculate its magnitude later. We have

∆xmax = v0x × ttot

Substituting the known values, we obtain

4m = v0x × 1.78 s
v0x = 4 m/1.78 s= 2.25 m/s

Therefore, applying the known equation

|v0 | = √v0x2 + v20y

we obtain for the magnitude of the initial velocity:

|v0 | = √802 + 2.252
= √80 + 6.25
= √86.25
= 9.29 m/s

The initial angle θ is calculated using the concept of tangent of the angle θ formed by the initial velocity vector and the horizontal direction. We have

tan θ = v0y/v0x = 8.94/2.25 = 3.97

Hence,

θ = arctan 3.97 = 75.90

At t = 0.3, we have

x(t) = v0x × t

and

y(t) = y0 + voy × t + g × t2/2

Substituting the known values, we obtain

x(0.3) = 2.25 × 0.3
= 0.675m

and

y(0.3) = 0 + 8.94 × 0.3 + (-10) × 0.32/2
= 2.682m - 0.450m
= 2.232m

Hence, at t = 0.3s the object is at (0.675m, 2.232m). This means it is 0.675 m on the right and 2.232 m above the starting point.

As stated earlier, the object is in two instants at the same height during a projectile. Therefore, we must use the quadratic equation of motion

y(t) = y0 + voy × t + g × t2/2

to calculate the required instants t. Thus, substituting the values, we obtain

3 = 0 + 8.94 × t + (-10) × t2/2
-5 × t2 + 8.94 × t - 3 = 0

Dividing both sides by (-5), we obtain

t2 - 1.788 × t + 0.6 = 0

Using any of the methods for solving quadratic equations, we obtain

t1 = 0.45s and t2 = 1.34s

Now, let's see what happens if an object thrown at an angle θ to the horizontal direction does not end its motion at the same level (height) but above or below the starting point. To illustrate this point we have used the numerical example below:

Example 2

A ball is kicked at 15 m/s from the terrace of a 12 m high building. The angle of the initial velocity with the horizon is 370. Ignore the air resistance and take the magnitude of g equal to 10 m/s2. Also, take cos 370 = 0.8 and sin 370 = 0.6.

Calculate:

1. The maximum height (from the ground) the object can reach.
2. What is the total time of flight?
3. How many metres away from the building is the ball when it touches the ground?
4. What is the ball's velocity just before touching the ground?

Solution 2

The figure below helps creating a clearer idea about the motion described in the problem.

To calculate the maximum height hmax, first we have to calculate the extra height h1 above the building and then add it with the building's height h0. We have:

v2yA - v20y = 2 × g × h1

We know that

v0x = v0 × cos θ = 15 × 0.8 = 12 m/s
v0y = v0 × sin θ = 15 × 0.6 = 9 m/s
vyA = 0

and

g = -10 m/s2

Therefore, we obtain after the substitutions:

02 - 92 = 2 × (-10) × h1
-81 = -20 × h1
h1 = -81/-20 = 4.05 m

Hence, the maximum height the object can reach is

hmax = h0 + h1
= 12m + 4.05m
= 16.05m

We can use the equation of projectile

y(t) = y0 + voy × t + g × t2/2

for calculating the total time of flight t. Thus, giving that at the end of motion the position of the object will be y = 0, we obtain after substituting the known values,

0 = 12 + 9 × t + (-10) × t2/2
-5 × t2 + 9 × t + 12 = 0
5 × t2-9 × t - 12 = 0

We have a quadratic equation (with variable t) which has its constants: A = 5, B = -9, C = -12. Therefore, we have

∆ = B2 - 4 × A × C
= (-9)2 - 4 × 5 × (-12)
= 81 + 240
= 321

Therefore, we have

t1 = -B - √/2 × A
= - (-9) - √321/2 × 5
= 9 - 17.9/10
= -8.9/10
= -0.89 s

This result is negative. But the time cannot be negative, so this value is automatically rejected. The second solution is

t2 = -B + √/2 × A
= -(-9) + √321/2 × 5
= 9 + 17.9/10
= 26.9/10
= 2.69 s

Hence,

ttot = t2 = 2.69 s

"How many metres away from the building is the ball when it touches the ground?" means: "calculate the total horizontal displacement ∆x of the object."

Therefore, we have:

∆x = v0x × ttot
= 12 m/s × 2.69 s
= 32.28 m

To calculate the magnitude of the object's final velocity just before touching the ground, we need its components vx and vy. One of them (vx) is already known as vx = v0x = 12 m/s. As for the vertical component of the final velocity vy, we have

v2y - v20y = 2 × g × ∆y

In our example ∆y = -12m as at the end of motion, the object is 12 m below the original height. Also, we take g = - 10 m/s2 as initially it was so. Therefore, we obtain after substitutions:

v2y-92 = 2 × (-10) × (-12)
v2y - 81 = 240
v2y = 240 + 81 = 321
vy = √321
= 17.9 m/s

Now, using the equation

|v| = √vx2 + v2y

we can calculate the magnitude of the final velocity when the object is just touching the ground. Thus,

|v| = √122 + 17.92
= √144 + 321
= √465
≈ 21.6 m/s

Whats next?

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