Projectile Motion
Independence of Motion
Projectile motion is a two-dimensional motion where horizontal and vertical components act independently under the sole influence of gravity.
$$x = v_x \times t$$
The horizontal distance depends only on constant horizontal velocity and time.
$x$=Horizontal displacement(m)
$v_x$=Horizontal velocity component (constant)(m/s)
$t$=Time elapsed(s)
No air resistance
→$v_x$ remains constant throughout the flight — the projectile has zero horizontal acceleration.
Horizontal Component: No horizontal force acts on the projectile (ignoring air resistance), so $a_x = 0$ and horizontal velocity remains constant.
Vertical Component: The object accelerates downwards at $g = 9.8$ m/s² regardless of its horizontal speed.
Stroboscopic Proof: A dropped ball and a horizontally projected ball hit the ground at the same time if released from the same height — their vertical motions are identical.
Dimensional Check: $[x] = [v_x][t] = (\text{m/s})(\text{s}) = \text{m}$ confirms displacement units.
The vertical displacement of any projectile follows the same equation as free fall, with the initial vertical velocity determining whether it rises first or falls directly.
$$y = v_{iy}t - \frac{1}{2}gt^2$$
Vertical position as a function of time, combining initial upward velocity and gravitational pull.
$y$=Vertical displacement (positive upward)(m)
$v_{iy}$=Initial vertical velocity ($v_i \sin\theta$)(m/s)
$g$=Acceleration due to gravity(m/s²)
Horizontal launch ($\theta = 0$)
→$v_{iy} = 0$, so $y = -\frac{1}{2}gt^2$ — pure free fall.
At time of flight ($y = 0$ again)
→The projectile returns to launch height, giving $t = \frac{2v_{iy}}{g}$.
Horizontal Launch: When thrown horizontally from a cliff, $v_{iy} = 0$ and the fall time depends only on height: $t = \sqrt{\frac{2h}{g}}$.
Sign Convention: Taking upward as positive, gravity is $-g$, so the vertical velocity decreases on the way up and increases (in magnitude) on the way down.
Oblique Projectile Components
When launched at an angle $\theta$, the initial velocity $v_i$ must be resolved into its orthogonal components using trigonometry.
$$v_{fy} = v_i \sin \theta - gt$$
The vertical velocity changes linearly over time due to constant gravitational acceleration.
$v_{fy}$=Vertical velocity at time $t$(m/s)
$v_i \sin \theta$=Initial vertical component of velocity(m/s)
$g$=Acceleration due to gravity(m/s²)
At peak height ($v_{fy} = 0$)
→Time to peak: $t_{peak} = \frac{v_i \sin\theta}{g}$, which is exactly half the total flight time.
At landing (same level)
→$v_{fy} = -v_i \sin\theta$ — equal magnitude but opposite direction to launch.
Resolution of Vectors: $v_{ix} = v_i \cos \theta$ (constant throughout) and $v_{iy} = v_i \sin \theta$ (changes due to gravity).
Constant Velocity: $v_x$ remains $v_i \cos \theta$ at every point on the parabolic trajectory.
Symmetry: The time taken to reach the peak is exactly half the total time of flight for ground-to-ground launches.
Speed at Landing: For same-level launches, the landing speed equals the launch speed: $v_f = v_i$. Only the direction changes.
Eliminating time from the horizontal and vertical displacement equations yields the trajectory equation — a parabola relating $y$ directly to $x$.
$$y = x \tan\theta - \frac{g x^2}{2 v_i^2 \cos^2\theta}$$
The path equation showing vertical position as a function of horizontal position — a downward-opening parabola.
$x \tan\theta$=The straight-line path the projectile would follow without gravity
$\frac{g x^2}{2 v_i^2 \cos^2\theta}$=The gravitational 'drop' below the straight-line path
$\theta = 45^\circ$
→$\tan\theta = 1$ and $\cos^2\theta = 0.5$, giving $y = x - \frac{gx^2}{v_i^2}$.
Derivation: Substitute $t = \frac{x}{v_i \cos\theta}$ from the horizontal equation into the vertical equation $y = v_i \sin\theta \cdot t - \frac{1}{2}gt^2$.
Parabolic Shape: The $x^2$ term confirms the trajectory is a parabola opening downward.
Dimensional Verification: $[x \tan\theta] = \text{m}$ and $\left[\frac{g x^2}{v_i^2 \cos^2\theta}\right] = \frac{(\text{m/s}^2)(\text{m}^2)}{(\text{m/s})^2} = \text{m}$ — both terms have correct dimensions.
Key Parameters: Height, Time, and Range
The maximum height is reached when the vertical velocity component becomes zero at the peak of the trajectory.
$$H = \frac{v_i^2 \sin^2 \theta}{2g}$$
Maximum vertical displacement, derived from $v_{fy}^2 = v_{iy}^2 - 2gH$ with $v_{fy} = 0$.
$H$=Maximum height above launch point(m)
$v_i \sin\theta$=Initial vertical velocity component(m/s)
$g$=Gravitational acceleration(m/s²)
$\theta = 90^\circ$ (vertical throw)
→$H = \frac{v_i^2}{2g}$ — maximum possible height for a given speed.
$\theta = 45^\circ$
→$H = \frac{v_i^2}{4g}$ and the range is $R = 4H$.
Proportionality Shortcut: $H \propto v_i^2$ — if speed triples, height increases by factor 9.
Limiting Case: At $\theta = 0^\circ$, $H = 0$ — a horizontally launched projectile gains no height above its launch point.
The time of flight is the total duration from launch to landing at the same elevation, and equals twice the time to reach peak height.
$$T = \frac{2 v_i \sin \theta}{g}$$
Total flight time for a ground-to-ground projectile, derived from the symmetry of the parabolic path.
$T$=Total time of flight(s)
$v_i \sin\theta$=Initial vertical velocity component(m/s)
$g$=Gravitational acceleration(m/s²)
$\theta = 90^\circ$
→$T = \frac{2v_i}{g}$ — maximum flight time (straight up and down).
$\theta = 30^\circ$ vs $\theta = 60^\circ$
→The $60^\circ$ projectile stays airborne longer ($\sin 60 > \sin 30$) even though both have the same range.
Proportionality Shortcut: $T \propto v_i$ (linear) — unlike $H$ and $R$ which scale as $v_i^2$.
Half-time to Peak: Time to reach maximum height is $t_{peak} = T/2 = \frac{v_i \sin\theta}{g}$.
Dimensional Check: $[T] = \frac{(\text{m/s})}{(\text{m/s}^2)} = \text{s}$ — dimensions are consistent.
The horizontal range of a projectile depends on both the initial speed and the launch angle through the double-angle sine function.
$$R = \frac{v_i^2 \sin 2\theta}{g}$$
The horizontal distance covered during the entire flight, found by multiplying $v_x$ by $T$.
$R$=Horizontal range(m)
$v_i$=Initial launch speed(m/s)
$\sin 2\theta$=Sine of double the launch angle
$\theta = 45^\circ$
→$\sin 90^\circ = 1$ gives maximum range: $R_{max} = \frac{v_i^2}{g}$.
Complementary angles ($\theta$ and $90^\circ - \theta$)
→$\sin 2\theta = \sin(180^\circ - 2\theta)$, so both angles give the same range.
Maximum Range: At $\theta = 45^\circ$, the range is maximized because $\sin 90^\circ = 1$.
Complementary Angle Rule: Angles like $30^\circ$ and $60^\circ$ produce identical ranges — the steeper angle gives more height but less horizontal speed, and vice versa.
Height-Range Ratio: At $\theta = 45^\circ$, $R = 4H$. This shortcut avoids recalculating from scratch.
Ground-to-Ground Only: This formula assumes launch and landing occur at the same vertical level.
Instantaneous Velocity
At any point during flight, the projectile's speed is found by combining its constant horizontal and changing vertical velocity components using the Pythagorean theorem.
$$V = \sqrt{v_{fx}^2 + v_{fy}^2}$$
The magnitude of the resultant velocity vector at any instant.
$V$=Instantaneous speed (magnitude of velocity)(m/s)
$v_{fx}$=Horizontal velocity ($= v_i \cos\theta$, constant)(m/s)
$v_{fy}$=Vertical velocity at time $t$ ($= v_i \sin\theta - gt$)(m/s)
At the peak
→$v_{fy} = 0$, so $V_{min} = v_{fx} = v_i \cos\theta$ — the minimum speed during flight.
At launch and landing (same level)
→$V = v_i$ — the speed is the same at both points (only direction changes).
Minimum Speed: Occurs at the peak where only horizontal velocity remains: $V_{min} = v_i \cos\theta$.
Direction of Velocity: The angle $\phi$ with the horizontal at any time is $\tan\phi = \frac{v_{fy}}{v_{fx}}$.
Velocity vs Speed: At the peak, velocity is purely horizontal (not zero!) — only the vertical component is zero.
Real-world Ballistics
Real-world projectiles, such as a ballistic missile, deviate from ideal parabolic paths due to aerodynamic forces.
$$R_{max} = \frac{v_i^2}{g}$$
The theoretical maximum range in a vacuum at $45^\circ$ — an upper bound for real-world performance.
$R_{max}$=Maximum possible range (vacuum, flat Earth)(m)
$v_i$=Initial launch speed(m/s)
Air friction present
→The actual range is significantly less than $R_{max}$ and the optimal angle is less than $45^\circ$.
Ideal vs Actual: Air drag causes the path to be shorter and asymmetric — the descent is steeper than the ascent.
Aerodynamic Forces: At high speeds, air friction can exceed gravitational force, making ideal equations completely unrealistic.
Ballistic Flight: An unpowered, unguided flight moving solely due to inertia and gravity.
Application Strategy
Worked Examples
Problem Statement
A ball is kicked at 30 m/s at an angle of $30^\circ$. Find the time of flight, max height, and range. (Take $g = 9.8$ m/s²)
Time of FlightMax HeightRange
Strategic Toolkit
$$T = \frac{2 v_i \sin \theta}{g}, \quad H = \frac{v_i^2 \sin^2\theta}{2g}, \quad R = \frac{v_i^2 \sin 2\theta}{g}$$
Step 1
Identify Givens
Extract variables: $v_i = 30$ m/s, $\theta = 30^\circ$, $g = 9.8$ m/s².
$v_i = 30, \theta = 30^\circ$
Step 2
Calculate Time of Flight
Substitute into the time formula.
$T = \frac{2 \times 30 \times \sin(30^\circ)}{9.8} = \frac{60 \times 0.5}{9.8} \approx 3.06$ s
Step 3
Calculate Max Height
Use the height formula.
$H = \frac{30^2 \times (\sin 30^\circ)^2}{2 \times 9.8} = \frac{900 \times 0.25}{19.6} \approx 11.5$ m
Step 4
Calculate Range
Use the range formula with double angle.
$R = \frac{30^2 \times \sin(60^\circ)}{9.8} = \frac{900 \times 0.866}{9.8} \approx 79.5$ m
Problem Statement
A stone is thrown horizontally from a 10m high cliff at 21 m/s. How far from the base does it land, and with what speed?
Horizontal LaunchIndependence of MotionInstantaneous Velocity
Strategic Toolkit
$$y = \frac{1}{2}gt^2, \quad V = \sqrt{v_x^2 + v_y^2}$$
Step 1
Vertical Fall Time
Since initial vertical velocity is zero ($\theta = 0$), solve for $t$ from the height equation.
$10 = \frac{1}{2}(9.8)t^2 \implies t = \sqrt{\frac{20}{9.8}} \approx 1.43$ s
Step 2
Horizontal Distance
Multiply horizontal velocity by the time in the air.
$x = v_x \times t = 21 \times 1.43 \approx 30$ m
Step 3
Find Vertical Velocity at Impact
The vertical velocity increases from zero due to gravity:
$v_y = gt = 9.8 \times 1.43 = 14$ m/s
Step 4
Calculate Landing Speed
Combine horizontal and vertical components using the Pythagorean theorem:
$V = \sqrt{v_x^2 + v_y^2} = \sqrt{21^2 + 14^2} = \sqrt{441 + 196} = \sqrt{637} \approx 25.2$ m/s
Problem Statement
A ball is thrown at 30 m/s. Compare the range and height at $45^\circ$ versus $60^\circ$. (Take $g = 9.8$ m/s²)
Complementary AnglesRangeHeight
Strategic Toolkit
$$R = \frac{v_i^2 \sin 2\theta}{g}, \quad H = \frac{v_i^2 \sin^2\theta}{2g}$$
Step 1
Range at 45°
Maximum range since $\sin 90^\circ = 1$.
$R_{45} = \frac{30^2}{9.8} = \frac{900}{9.8} \approx 91.8$ m
Step 2
Height at 45°
Use the height formula with $\sin^2 45^\circ = 0.5$.
$H_{45} = \frac{900 \times 0.5}{19.6} \approx 23$ m
Step 3
Range at 60°
Note $60^\circ$ and $30^\circ$ are complementary, so $R_{60} = R_{30}$.
$R_{60} = \frac{900 \times \sin 120^\circ}{9.8} = \frac{900 \times 0.866}{9.8} \approx 79.5$ m
Step 4
Height at 60°
The steeper angle gives significantly more height.
$H_{60} = \frac{900 \times (\sin 60^\circ)^2}{19.6} = \frac{900 \times 0.75}{19.6} \approx 34.4$ m
Step 5
Key Insight
The $60^\circ$ launch gains 50% more height than $45^\circ$ but loses about 13% of the maximum range. Complementary pair ($30^\circ$ and $60^\circ$) would have the same range of 79.5 m.
$R_{30} = R_{60} = 79.5$ m, but $H_{60} = 34.4$ m while $H_{30} = 11.5$ m
Memorize formulas
Time of Flight
Total time in air for a ground-to-ground projectile.
Formula
$T = \frac{2 v_i \sin \theta}{g}$
Maximum Height
Peak vertical displacement reached by the projectile.
Formula
$H = \frac{v_i^2 \sin^2 \theta}{2g}$
Horizontal Range
Total horizontal distance covered. Uses double angle $\sin 2\theta$.
Formula
$R = \frac{v_i^2 \sin 2\theta}{g}$
Trajectory Equation
Path equation relating $y$ to $x$ directly (parabola).
Formula
$y = x \tan\theta - \frac{g x^2}{2 v_i^2 \cos^2\theta}$
Instantaneous Velocity
Speed at any point during flight.
Formula
$V = \sqrt{v_x^2 + v_y^2}$
Review
1
A bomber drops a bomb while flying horizontally at a constant speed. Ignoring air resistance, where will the bomber be when the bomb hits the ground?
Ans
Directly above the impact point
Review
Tap to reveal
Match the horizontal motion
When the bomb is released, it already has the bomber's horizontal speed, and with no air resistance there is no horizontal acceleration to change that speed.
The bomb and bomber cover the same horizontal distance in the same time.
Compare positions at impact
Because their horizontal motions stay synchronized while the bomb falls, the bomber remains vertically aligned with the bomb's landing point.
The bomber is directly above the impact point.
2
Calculate the range of a projectile launched at $15^\circ$ if its range at $45^\circ$ is 100m (same initial velocity).
Ans
$50$ m
Review
Tap to reveal
Use the range proportionality
For the same launch speed, range varies as $\sin 2\theta$, so we only need the ratio of the sine terms.
$$\frac{R_{15}}{R_{45}} = \frac{\sin 30^\circ}{\sin 90^\circ} = \frac{1/2}{1}$$
Scale from the known range
Multiply the 45-degree range by this factor to get the new range.
$$R_{15} = 100 \times \frac{1}{2} = 50\ \text{m}$$
3
A projectile launched at $40^\circ$ with speed $v_i$ has a range $R$. At what other angle (with the same $v_i$) will it land at a distance $R/2$?
Ans
$15^\circ$
Review
Tap to reveal
Write the range condition
For the same initial speed, range is proportional to $\sin 2\theta$. Halving the range means the new angle must satisfy half the original sine value.
$$\sin 2\alpha = \frac{1}{2}\sin 80^\circ$$
Estimate the angle
Since $\sin 80^\circ \approx 0.985$, the target is about $0.492$. That corresponds to $2\alpha \approx 30^\circ$.
$$\alpha \approx 15^\circ$$
4
A projectile is fired at $60^\circ$ with speed $v_i$. What is the ratio of its maximum height to its range ($H/R$)?
Ans
$\frac{\sqrt{3}}{4}$
Review
Tap to reveal
Substitute the launch angle into H and R
Use the standard projectile formulas for maximum height and range, then replace $\theta$ with $60^\circ$.
$$H = \frac{v_i^2\sin^2 60^\circ}{2g},\quad R = \frac{v_i^2\sin 120^\circ}{g}$$
Form the ratio and simplify
Cancel the common factor $v_i^2/g$ and reduce the remaining trig values.
$$\frac{H}{R} = \frac{\sin^2 60^\circ}{2\sin 120^\circ} = \frac{\sqrt{3}}{4}$$
5
What is the angle between the velocity vector and the acceleration vector at the highest point of a projectile's path?
Ans
$90^\circ$
Review
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Identify the two directions at the peak
At the highest point, the vertical component of velocity is zero, so velocity is horizontal. Gravity still acts downward, so acceleration is vertical.
Velocity is horizontal and acceleration is downward vertical.
Read off the angle
Horizontal and vertical directions are perpendicular.
$$90^\circ$$
6
A ball is thrown horizontally at 20 m/s from a tower of height 80m. How long does it take to hit the ground? (Take $g = 10$ m/s²)
Ans
$4$ s
Review
Tap to reveal
Use vertical free-fall motion
A horizontal launch has zero initial vertical velocity, so the vertical drop is governed only by gravity.
$$h = \tfrac{1}{2}gt^2$$
Solve for time
Substitute $h = 80$ m and $g = 10$ m/s$^2$ and isolate $t$.
$$80 = \tfrac{1}{2}(10)t^2 \Rightarrow t^2 = 16 \Rightarrow t = 4\ \text{s}$$
7
If the initial velocity of a projectile is tripled while keeping the angle fixed, by what factor does the range change?
Ans
$9$
Review
Tap to reveal
Use the velocity-squared dependence
With launch angle fixed, the sine term and gravity are unchanged, so the range depends only on the square of the initial speed.
$$R \propto v_i^2$$
Apply the tripling factor
Tripling the speed multiplies the range by $3^2$.
$$R' = 9R$$
8
A projectile launched at $30^\circ$ achieves a horizontal range of 100m. What is its maximum height? (Take $g = 10$ m/s²)
Ans
$14.4$ m
Review
Tap to reveal
Use the height-to-range relation
For the same launch speed, combining the standard projectile formulas gives $H/R = \tan\theta/4$.
$$\frac{H}{R} = \frac{\tan\theta}{4}$$
Substitute the given angle and range
Insert $\theta = 30^\circ$ and $R = 100$ m, then evaluate the tangent.
$$H = 100 \cdot \frac{\tan 30^\circ}{4} = \frac{100}{4\sqrt{3}} \approx 14.4\ \text{m}$$
9
At what point in its trajectory does a projectile have its minimum speed?
Ans
At the highest point
Review
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Check the velocity components
The horizontal component stays constant, but the vertical component decreases to zero at the top of the path.
$$v_y = 0 \text{ at the highest point}$$
Compute the speed there
With no vertical component, the speed is just the constant horizontal component, which is smaller than the launch or landing speed.
$$V_{\min} = v_x = v_i\cos\theta$$
10
A projectile is fired and lands at the same level. If $\theta = 30^\circ$ and $v_i = 50$ m/s, what is the vertical velocity just before hitting the ground? (Take $g = 10$ m/s²)
Ans
$25$ m/s downward
Review
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Use symmetry for same-level landing
For a projectile that lands at the same height, the vertical speed on the way down matches the launch vertical speed in magnitude.
$$|v_{fy}| = v_i\sin\theta$$
Assign the downward sign
Because the projectile is descending just before impact, the vertical component is negative.
$$v_{fy} = -50\sin 30^\circ = -25\ \text{m/s}$$
11
Which quantity remains constant throughout the flight of a projectile (ignoring air resistance)?
Ans
Horizontal velocity
Review
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Identify the unforced direction
With air resistance neglected, gravity acts only vertically, so there is no horizontal acceleration.
$$a_x = 0$$
Conclude the constant component
Zero horizontal acceleration means the horizontal component of velocity does not change during the motion.
$$v_x = v_i\cos\theta = \text{constant}$$
12
A projectile is launched at $76^\circ$ with $v_i = 40$ m/s. What is the ratio of its time of flight to the time it takes to reach maximum height? (Take $g = 10$ m/s²)
Ans
$2$
Review
Tap to reveal
Write the upward time
The time to reach the top is set by the initial vertical component divided by gravity.
$$t_{up} = \frac{v_i\sin\theta}{g}$$
Use symmetry of the path
For a projectile that lands at the same level, the descent takes the same time as the ascent, so total flight time is twice the upward time.
$$T = 2t_{up} \Rightarrow \frac{T}{t_{up}} = 2$$
Question 1 of 12
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1
A bomber drops a bomb while flying horizontally at a constant speed. Ignoring air resistance, where will the bomber be when the bomb hits the ground?