The Central Idea
A projectile is an object that, after launch, moves primarily under the influence of gravity.
In the standard idealized model:
- air resistance is neglected;
- gravity is constant;
- gravity acts vertically downward;
- horizontal acceleration is zero;
- horizontal and vertical motions share the same time parameter.
The key strategy is to analyze the horizontal and vertical components separately.
Choosing Coordinates
A common convention is:
- positive \(x\) points horizontally to the right;
- positive \(y\) points vertically upward.
With this convention, gravitational acceleration is
\[ \boxed{ \mathbf{a} = \langle0,-g\rangle }. \]Near Earth's surface,
\[ g\approx9.8\frac{\text{m}}{\text{s}^2}. \]Horizontal and Vertical Acceleration
Under the ideal projectile model,
\[ \boxed{ a_x=0 } \]and
\[ \boxed{ a_y=-g }. \]Therefore, the horizontal velocity is constant while the vertical velocity changes linearly with time.
Initial Velocity Components
Suppose a projectile is launched with initial speed \(v_0\) at angle \(\theta\) above the horizontal.
The horizontal component is
\[ \boxed{ v_{0x}=v_0\cos\theta } \]and the vertical component is
\[ \boxed{ v_{0y}=v_0\sin\theta }. \]Why We Resolve the Velocity into Components
Gravity acts only in the vertical direction in this model.
Therefore, the horizontal and vertical velocity components obey different equations:
\[ v_x(t)=v_{0x} \]while
\[ v_y(t)=v_{0y}-gt. \]Treating the components separately turns one curved motion into two simpler one-dimensional motions.
Horizontal Position
Since horizontal acceleration is zero, horizontal position changes at constant velocity.
If the projectile begins at horizontal position \(x_0\), then
\[ \boxed{ x(t) = x_0+v_{0x}t }. \]Using the launch-angle component,
\[ \boxed{ x(t) = x_0+v_0\cos\theta\,t }. \]Vertical Position
Vertical acceleration is constant:
\[ a_y=-g. \]Therefore, if the projectile begins at height \(y_0\),
\[ \boxed{ y(t) = y_0+v_{0y}t-\frac12gt^2 }. \]Using the launch-angle component,
\[ \boxed{ y(t) = y_0+v_0\sin\theta\,t-\frac12gt^2 }. \]The Position Vector
Combine the coordinate functions:
\[ \boxed{ \mathbf{r}(t) = \left\langle x_0+v_0\cos\theta\,t,\; y_0+v_0\sin\theta\,t-\frac12gt^2 \right\rangle }. \]This vector-valued function describes the complete ideal projectile motion.
The Velocity Vector
Differentiate the position vector:
\[ \boxed{ \mathbf{v}(t) = \left\langle v_0\cos\theta,\; v_0\sin\theta-gt \right\rangle }. \]The horizontal component remains constant while the vertical component decreases linearly.
The Acceleration Vector
Differentiate velocity:
\[ \boxed{ \mathbf{a}(t) = \langle0,-g\rangle }. \]The acceleration vector remains constant throughout the ideal motion.
Example: Launch Components
A projectile is launched at
\[ v_0=20\frac{\text{m}}{\text{s}} \]at angle
\[ \theta=30^\circ. \]Then
\[ v_{0x}=20\cos30^\circ \] \[ = 20\left(\frac{\sqrt3}{2}\right) \] \[ = \boxed{ 10\sqrt3\frac{\text{m}}{\text{s}} }. \]Also,
\[ v_{0y}=20\sin30^\circ \] \[ =20\left(\frac12\right) \] \[ = \boxed{ 10\frac{\text{m}}{\text{s}} }. \]Example: Position Functions
Suppose the same projectile begins at the origin.
Then
\[ x_0=0, \qquad y_0=0. \]Its position functions are
\[ \boxed{ x(t)=10\sqrt3\,t } \]and
\[ \boxed{ y(t)=10t-4.9t^2 }. \]The Projectile Path Is Parabolic
For nonvertical ideal projectile motion, eliminate \(t\).
From
\[ x=x_0+v_0\cos\theta\,t, \]we obtain
\[ t = \frac{x-x_0}{v_0\cos\theta}. \]Substitute this into the vertical equation.
Trajectory Equation
Substitution gives
\[ y = y_0 + v_0\sin\theta \left( \frac{x-x_0}{v_0\cos\theta} \right) - \frac12g \left( \frac{x-x_0}{v_0\cos\theta} \right)^2. \]Simplifying,
\[ \boxed{ y = y_0 + (x-x_0)\tan\theta - \frac{ g(x-x_0)^2 }{ 2v_0^2\cos^2\theta } }. \]This is quadratic in \(x\), so the ideal trajectory is a parabola.
Horizontal Velocity
In ideal projectile motion,
\[ \boxed{ v_x(t)=v_0\cos\theta }. \]It remains constant because
\[ a_x=0. \]Vertical Velocity
The vertical velocity is
\[ \boxed{ v_y(t)=v_0\sin\theta-gt }. \]It begins positive for an upward launch, decreases to zero at the top, and becomes negative during descent.
Speed during Flight
Speed is the magnitude of the velocity vector:
\[ \boxed{ v(t) = \sqrt{ v_x(t)^2+v_y(t)^2 } }. \]Therefore,
\[ \boxed{ v(t) = \sqrt{ v_0^2\cos^2\theta + \left( v_0\sin\theta-gt \right)^2 } }. \]The Highest Point
At the highest point, the vertical component of velocity is zero:
\[ \boxed{ v_y=0 }. \]But the horizontal component usually remains nonzero.
Therefore, the projectile is generally still moving at the top.
Time to Maximum Height
Set
\[ v_y(t)=0. \]Then
\[ v_0\sin\theta-gt=0. \]Therefore,
\[ \boxed{ t_{\text{top}} = \frac{v_0\sin\theta}{g} }. \]Maximum Height above the Launch Point
Substitute
\[ t_{\text{top}} = \frac{v_0\sin\theta}{g} \]into the vertical displacement equation:
\[ \Delta y = v_0\sin\theta\,t - \frac12gt^2. \]Simplifying gives
\[ \boxed{ H = \frac{ v_0^2\sin^2\theta }{ 2g } }. \]Maximum Absolute Height
If the projectile begins at height \(y_0\), then its maximum vertical coordinate is
\[ \boxed{ y_{\max} = y_0 + \frac{ v_0^2\sin^2\theta }{ 2g } }. \]Example: Time to the Top
A projectile has initial vertical velocity
\[ v_{0y}=19.6\frac{\text{m}}{\text{s}}. \]Then
\[ t_{\text{top}} = \frac{19.6}{9.8} \] \[ = \boxed{2\text{ s}}. \]Example: Maximum Height
Using
\[ v_{0y}=19.6\frac{\text{m}}{\text{s}}, \]the height gained is
\[ H = \frac{19.6^2}{2(9.8)}. \]Since
\[ 19.6^2=384.16, \]we obtain
\[ H=\frac{384.16}{19.6} \] \[ = \boxed{19.6\text{ m}}. \]Time of Flight for Equal Launch and Landing Heights
Suppose
\[ y_0=y_f. \]Then
\[ 0 = v_0\sin\theta\,t - \frac12gt^2. \]Factor:
\[ t \left( v_0\sin\theta-\frac12gt \right) = 0. \]The nonzero flight time is
\[ \boxed{ T = \frac{ 2v_0\sin\theta }{ g } }. \]Symmetry of Equal-Height Projectile Motion
When launch and landing heights are equal and air resistance is neglected, the motion is symmetric in time.
Therefore,
\[ \boxed{ T=2t_{\text{top}} }. \]The time spent rising equals the time spent falling back to the launch height.
Range for Equal Launch and Landing Heights
Horizontal range is horizontal speed multiplied by flight time:
\[ R = v_0\cos\theta \left( \frac{2v_0\sin\theta}{g} \right). \]Therefore,
\[ R = \frac{ 2v_0^2\sin\theta\cos\theta }{ g }. \]Using
\[ 2\sin\theta\cos\theta=\sin2\theta, \]we obtain
\[ \boxed{ R = \frac{ v_0^2\sin2\theta }{ g } }. \]Maximum Range on Level Ground
For fixed launch speed \(v_0\), the level-ground range is
\[ R = \frac{v_0^2}{g}\sin2\theta. \]The largest possible value of
\[ \sin2\theta \]is \(1\).
Therefore,
\[ 2\theta=90^\circ \]and
\[ \boxed{\theta=45^\circ}. \]This result assumes equal launch and landing heights and no air resistance.
Complementary Launch Angles
Since
\[ \sin2\theta = \sin(180^\circ-2\theta), \]complementary launch angles
\[ \theta \]and
\[ 90^\circ-\theta \]produce the same range under the ideal equal-height model.
Example: Complementary Angles
Launch angles
\[ 30^\circ \]and
\[ 60^\circ \]are complementary.
Therefore, for the same launch speed and equal launch and landing heights, they have the same ideal range.
The \(60^\circ\) trajectory reaches a greater height and remains airborne longer.
Example: Level-Ground Time of Flight
A projectile is launched at
\[ v_0=30\frac{\text{m}}{\text{s}} \]and
\[ \theta=30^\circ. \]The initial vertical component is
\[ v_{0y}=30\sin30^\circ=15. \]Therefore,
\[ T = \frac{2(15)}{9.8} \] \[ = \boxed{ \frac{30}{9.8}\text{ s} }. \]Example: Level-Ground Range
For the same projectile,
\[ R = \frac{ 30^2\sin60^\circ }{ 9.8 }. \]Therefore,
\[ \boxed{ R = \frac{ 900(\sqrt3/2) }{ 9.8 } }. \]This exact expression may be approximated numerically when desired.
Unequal Launch and Landing Heights
If the projectile lands at a different height from where it was launched, the simple symmetric formulas for total flight time and range do not generally apply.
Instead, solve
\[ \boxed{ y_f = y_0+v_{0y}t-\frac12gt^2 }. \]This usually produces a quadratic equation in \(t\).
Example: Horizontal Launch from a Height
Suppose an object is launched horizontally from height \(h\) with speed \(v_0\).
Then
\[ v_{0x}=v_0 \]and
\[ v_{0y}=0. \]The motion equations become
\[ \boxed{ x(t)=x_0+v_0t } \]and
\[ \boxed{ y(t)=h-\frac12gt^2 }. \]Time to Fall from Height \(h\)
If the ground is
\[ y=0, \]then
\[ 0=h-\frac12gt^2. \]Therefore,
\[ \frac12gt^2=h \]and
\[ \boxed{ t = \sqrt{\frac{2h}{g}} }. \]Horizontal Range from a Height
For a horizontal launch,
\[ x-x_0=v_0t. \]Using
\[ t = \sqrt{\frac{2h}{g}}, \]the horizontal distance traveled before impact is
\[ \boxed{ R = v_0 \sqrt{\frac{2h}{g}} }. \]Example: Horizontal Launch
An object is launched horizontally at
\[ 12\frac{\text{m}}{\text{s}} \]from a height of
\[ 19.6\text{ m}. \]Time to fall:
\[ t = \sqrt{ \frac{2(19.6)}{9.8} } \] \[ = \sqrt4 \] \[ = \boxed{2\text{ s}}. \]Horizontal distance:
\[ R=12(2) \] \[ = \boxed{24\text{ m}}. \]Vertical Launch
A purely vertical launch is the special case
\[ v_{0x}=0. \]Then
\[ x(t)=x_0 \]throughout the motion.
The projectile reduces to one-dimensional vertical motion:
\[ y(t) = y_0+v_0t-\frac12gt^2. \]Velocity at the Top
At maximum height,
\[ v_y=0. \]For a nonvertical launch,
\[ v_x=v_0\cos\theta. \]Therefore, the speed at the top is
\[ \boxed{ v_{\text{top}} = |v_0\cos\theta| }. \]Acceleration at the Top
Even though
\[ v_y=0 \]at the top, gravity does not disappear.
The acceleration remains
\[ \boxed{ \mathbf{a} = \langle0,-g\rangle }. \]Zero vertical velocity does not imply zero acceleration.
Velocity on the Way Down
After the top,
\[ v_y<0 \]under the upward-positive convention.
The horizontal component remains unchanged:
\[ v_x=v_0\cos\theta. \]Thus the velocity vector tilts progressively downward.
Equal Heights and Speed Symmetry
In the ideal model, when a projectile returns to the same height from which it was launched:
- the horizontal velocity component is unchanged;
- the vertical velocity component has the same magnitude but opposite sign;
- the speed equals the launch speed.
Why the Speed Matches at Equal Heights
At launch,
\[ \mathbf{v}_0 = \langle v_{0x}, v_{0y} \rangle. \]On returning to the same height,
\[ \mathbf{v} = \langle v_{0x}, -v_{0y} \rangle. \]Both have magnitude
\[ \sqrt{ v_{0x}^2+v_{0y}^2 }. \]Therefore,
\[ \boxed{v=v_0}. \]Projectile Motion as Constant Vector Acceleration
Projectile motion is an important example of the vector equation
\[ \boxed{ \mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}_0t + \frac12\mathbf{a}t^2 }. \]Here
\[ \mathbf{a} = \langle0,-g\rangle. \]Expanding the vector equation produces the horizontal and vertical component formulas.
Vector Form with an Initial Angle
The initial velocity vector is
\[ \mathbf{v}_0 = \langle v_0\cos\theta, v_0\sin\theta \rangle. \]Therefore,
\[ \mathbf{r}(t) = \langle x_0,y_0\rangle + t \langle v_0\cos\theta, v_0\sin\theta \rangle + \frac12t^2 \langle0,-g\rangle. \]Finding the Launch Angle from Velocity Components
If the initial velocity components are known,
\[ v_{0x} \]and
\[ v_{0y}, \]then
\[ \tan\theta = \frac{v_{0y}}{v_{0x}}. \]Thus
\[ \boxed{ \theta = \tan^{-1} \left( \frac{v_{0y}}{v_{0x}} \right) } \]with quadrant interpretation as needed.
Finding Initial Speed from Components
The initial speed is the magnitude of the initial velocity vector:
\[ \boxed{ v_0 = \sqrt{ v_{0x}^2+v_{0y}^2 } }. \]Example: Recovering Speed and Angle
Suppose
\[ v_{0x}=12\frac{\text{m}}{\text{s}} \]and
\[ v_{0y}=5\frac{\text{m}}{\text{s}}. \]Then
\[ v_0 = \sqrt{12^2+5^2} \] \[ = \boxed{ 13\frac{\text{m}}{\text{s}} }. \]The launch angle satisfies
\[ \boxed{ \tan\theta=\frac5{12} }. \]Projectile Motion and Graphs
The coordinate functions have different graph shapes:
- \(x(t)\) is linear when horizontal acceleration is zero;
- \(y(t)\) is quadratic because vertical acceleration is constant;
- \(v_x(t)\) is constant;
- \(v_y(t)\) is linear;
- \(a_x(t)=0\);
- \(a_y(t)=-g\).
A later lesson on position, velocity, and acceleration graphs will organize these relationships more systematically.
Common Error: Treating Horizontal Velocity as Changing because the Path Curves
In the ideal model,
\[ a_x=0. \]Therefore,
\[ \boxed{ v_x=\text{constant} }. \]The path curves because the vertical velocity changes, not because the horizontal velocity changes.
Common Error: Setting Acceleration to Zero at the Top
At maximum height,
\[ v_y=0. \]But
\[ a_y=-g. \]Gravity continues acting throughout the ideal flight.
Common Error: Setting the Entire Velocity to Zero at the Top
At the top of a nonvertical projectile trajectory, only the vertical component is zero:
\[ v_y=0. \]Usually,
\[ v_x\ne0. \]Therefore, the projectile is still moving horizontally.
Common Error: Using \(g\) with the Wrong Sign
If upward is positive, then
\[ a_y=-g. \]If downward is chosen positive instead, then gravity is positive.
The sign must agree with the coordinate convention.
Common Error: Mixing Horizontal and Vertical Equations
Horizontal quantities belong in the horizontal model:
\[ x=x_0+v_{0x}t. \]Vertical quantities belong in the vertical model:
\[ y=y_0+v_{0y}t-\frac12gt^2. \]The shared variable connecting them is time.
Common Error: Using the Level-Ground Range Formula for Unequal Heights
The formula
\[ R=\frac{v_0^2\sin2\theta}{g} \]assumes the projectile lands at the same height from which it was launched.
If the landing height differs, solve the vertical position equation for the actual flight time first.
Common Error: Assuming \(45^\circ\) Always Gives Maximum Range
The \(45^\circ\) maximum-range result depends on the ideal level-ground model.
Different launch and landing heights, air resistance, or other constraints can change the optimal angle.
Common Error: Forgetting to Check the Time Domain
Solving the vertical equation may produce two roots.
One root may represent the launch instant, while another represents landing.
Negative roots may lie outside the modeled time interval.
Always interpret roots physically.
A Reliable Procedure for Projectile Problems
- Choose coordinate directions.
- Identify \(x_0\) and \(y_0\).
- Resolve \(v_0\) into \(v_{0x}\) and \(v_{0y}\).
- Write \(x(t)\) and \(y(t)\).
- Write \(v_x(t)\) and \(v_y(t)\) if velocity is needed.
- Use the vertical equation to solve for important times.
- Use those times in the horizontal equation for horizontal positions or range.
- Check signs, units, and the allowed time interval.
A Reliable Procedure for Maximum Height
- Find the initial vertical velocity \(v_{0y}\).
- Set \(v_y(t)=0\).
- Solve for \(t_{\text{top}}\).
- Substitute into \(y(t)\).
- Distinguish height gained from absolute height above the chosen origin.
A Reliable Procedure for Range
- Determine the landing height.
- Solve the vertical position equation for the landing time.
- Select the physically relevant positive time.
- Substitute that time into \(x(t)\).
- Subtract the initial horizontal coordinate if range is requested.
Example: Full Projectile Workflow
A projectile is launched from the origin with
\[ v_0=25\frac{\text{m}}{\text{s}} \]at angle
\[ \theta=37^\circ. \]Its initial components are
\[ v_{0x}=25\cos37^\circ \]and
\[ v_{0y}=25\sin37^\circ. \]The position functions are
\[ \boxed{ x(t) = 25\cos37^\circ\,t } \]and
\[ \boxed{ y(t) = 25\sin37^\circ\,t - 4.9t^2 }. \]Time to maximum height:
\[ \boxed{ t_{\text{top}} = \frac{ 25\sin37^\circ }{ 9.8 } }. \]For equal launch and landing heights, total flight time is
\[ \boxed{ T = \frac{ 50\sin37^\circ }{ 9.8 } }. \]Practice 1: Resolve the Initial Velocity
A projectile is launched at \(20\text{ m/s}\) at \(60^\circ\). Find \(v_{0x}\) and \(v_{0y}\).
Practice 2: Position Equations
A projectile begins at the origin with
\[ v_{0x}=8\frac{\text{m}}{\text{s}} \]and
\[ v_{0y}=12\frac{\text{m}}{\text{s}}. \]Write \(x(t)\) and \(y(t)\).
Practice 3: Velocity Components
For the projectile in Practice 2, write \(v_x(t)\) and \(v_y(t)\).
Practice 4: Time to Maximum Height
A projectile has initial vertical velocity \(14.7\text{ m/s}\). Find the time to maximum height.
Practice 5: Maximum Height
A projectile is launched vertically upward from ground level at \(19.6\text{ m/s}\). Find its maximum height.
Practice 6: Level-Ground Flight Time
A projectile is launched from ground level at \(v_0=24\text{ m/s}\) and \(\theta=30^\circ\). Find its total ideal flight time.
Practice 7: Horizontal Launch
An object is launched horizontally at \(10\text{ m/s}\) from a height of \(44.1\text{ m}\). Find the time to reach the ground.
Practice 8: Horizontal Range from a Height
Continue Practice 7. Find the horizontal distance traveled before impact.
Practice 9: Position Vector
Write the position vector for a projectile launched from \((2,5)\) with initial velocity
\[ \langle6,10\rangle \frac{\text{m}}{\text{s}}. \]Practice 10: Identify the Error
A student says that at the highest point of a projectile's path, both velocity and acceleration are zero.
Explain the error.
Practice Answers
Practice 1
\[ v_{0x} = 20\cos60^\circ \] \[ = 20\left(\frac12\right) \] \[ = \boxed{ 10\frac{\text{m}}{\text{s}} }. \] \[ v_{0y} = 20\sin60^\circ \] \[ = 20\left(\frac{\sqrt3}{2}\right) \] \[ = \boxed{ 10\sqrt3\frac{\text{m}}{\text{s}} }. \]Practice 2
\[ \boxed{ x(t)=8t } \]and
\[ \boxed{ y(t)=12t-4.9t^2 }. \]Practice 3
\[ \boxed{ v_x(t)=8 } \]and
\[ \boxed{ v_y(t)=12-9.8t }. \]Practice 4
\[ t_{\text{top}} = \frac{14.7}{9.8} \] \[ = \boxed{1.5\text{ s}}. \]Practice 5
\[ H = \frac{19.6^2}{2(9.8)} \] \[ = \boxed{19.6\text{ m}}. \]Practice 6
The initial vertical component is
\[ v_{0y} = 24\sin30^\circ \] \[ = 12\frac{\text{m}}{\text{s}}. \]Therefore,
\[ T = \frac{2(12)}{9.8} \] \[ = \boxed{ \frac{24}{9.8}\text{ s} }. \]Practice 7
\[ t = \sqrt{ \frac{2h}{g} }. \] \[ = \sqrt{ \frac{2(44.1)}{9.8} }. \] \[ = \sqrt9 \] \[ = \boxed{3\text{ s}}. \]Practice 8
\[ R=v_0t \] \[ =10(3) \] \[ = \boxed{30\text{ m}}. \]Practice 9
Initial position:
\[ \mathbf{r}_0=\langle2,5\rangle. \]Initial velocity:
\[ \mathbf{v}_0=\langle6,10\rangle. \]Acceleration:
\[ \mathbf{a}=\langle0,-9.8\rangle. \]Therefore,
\[ \boxed{ \mathbf{r}(t) = \langle 2+6t,\; 5+10t-4.9t^2 \rangle }. \]Practice 10
At the highest point, only the vertical component of velocity is zero:
\[ v_y=0. \]For a nonvertical launch, the horizontal component remains nonzero:
\[ v_x=v_0\cos\theta. \]Gravity also continues acting, so
\[ \boxed{ \mathbf{a} = \langle0,-g\rangle }. \]Therefore, neither the full velocity nor the acceleration is generally zero at the top.
What to Remember
Resolve the initial velocity into components:
\[ \boxed{ v_{0x}=v_0\cos\theta, \qquad v_{0y}=v_0\sin\theta }. \]The ideal projectile equations are
\[ \boxed{ x(t) = x_0+v_0\cos\theta\,t } \] \[ \boxed{ y(t) = y_0+v_0\sin\theta\,t-\frac12gt^2 } \] \[ \boxed{ v_x(t)=v_0\cos\theta } \] \[ \boxed{ v_y(t)=v_0\sin\theta-gt } \] \[ \boxed{ \mathbf{a} = \langle0,-g\rangle }. \]For equal launch and landing heights,
\[ \boxed{ T = \frac{2v_0\sin\theta}{g} } \] \[ \boxed{ H = \frac{v_0^2\sin^2\theta}{2g} } \] \[ \boxed{ R = \frac{v_0^2\sin2\theta}{g} }. \]The most important conceptual facts are:
- horizontal and vertical motions share the same time;
- horizontal velocity is constant in the ideal model;
- vertical acceleration is constant and downward;
- the ideal trajectory is parabolic;
- at maximum height, vertical velocity is zero but acceleration is not;
- special range formulas must be used only when their assumptions hold.
Where We Go Next
Projectile motion was described using a particular coordinate system with a chosen origin and chosen horizontal and vertical directions. But the same physical motion can be described from different moving or stationary viewpoints.
The next lesson, Reference Frames, examines how position and velocity descriptions depend on the observer and how motion is translated between coordinate frames.