Projectile motion is the motion of an object thrown or projected into the air, subject only to the acceleration due to gravity. It moves along a curved path called a trajectory.
Basic Concepts
A projectile has both horizontal and vertical components of velocity. These components act independently:
Horizontal velocity \( v_x \) stays constant (no acceleration horizontally, ignoring air resistance).
Vertical velocity \( v_y \) changes due to gravity \( g \) acting downward.
Velocity Components
If a projectile is launched at an initial speed \( v_0 \) at an angle \( \theta \) above the horizontal:
Because no horizontal forces act on the projectile (assuming no air resistance), the horizontal velocity remains constant:
\( v_x = v_{0x} = v_0 \cos \theta \)
The horizontal position \( x \) changes linearly with time:
\( x = v_x t = v_0 \cos \theta \times t \)
Vertical Motion
Gravity causes a vertical acceleration downward of magnitude \( g \approx 9.8 \, m/s^2 \). The vertical velocity changes with time:
\( v_y = v_{0y} - g t = v_0 \sin \theta - g t \)
The vertical position \( y \) changes as:
\( y = v_{0y} t - \frac{1}{2} g t^2 = v_0 \sin \theta \times t - \frac{1}{2} g t^2 \)
Key Point: Horizontal Velocity Does NOT Affect Time of Flight
One important fact about projectile motion is that the time it takes for an object to hit the ground depends only on the vertical motion — specifically, how long it takes to fall from its initial height under gravity.
This means that no matter how fast the projectile moves horizontally, the time of flight is the same as if you simply dropped it straight down from the same initial height: y0.
Mathematically, the time to fall from height \( y_0 \) to ground level \( y = 0 \) (ignoring air resistance) is found by solving:
\[
0 = y_0 + v_{0y} t - \frac{1}{2} g t^2
\]
If the projectile starts at ground level with vertical launch velocity \( v_{0y} = v_0 \sin \theta \), the total time of flight until it returns to \( y=0 \) is:
\[
t_{flight} = \frac{2 v_0 \sin \theta}{g}
\]
You do NOT need to memorize this formula. Just remember the range of a projectile is the horizontal component of velocity times the time it takes to fall.
Notice this formula does not include the horizontal velocity \( v_x \). The horizontal speed does not change how long the projectile stays in the air.
Range of the Projectile
The horizontal distance (range) the projectile travels before hitting the ground is:
You do NOT need to memorize this formula. Just remember the range of a projectile is the horizontal component of velocity times the time it takes to fall.
Maximum Height
The highest point in the projectile's trajectory happens when the vertical velocity becomes zero:
\[
v_y = v_0 \sin \theta - g t = 0 \implies t_{peak} = \frac{v_0 \sin \theta}{g}
\]
You do NOT need to memorize this formula either. Just remember that at the max height of a projectile, the vertical velocity is 0 (if thrown above a 0 degree angle). Using this, you can use the formula in the 1st part to get the time.
Interactive Projectile Visualization
Additional Resources
Video Explanation: Time of Flight Independent of Horizontal Velocity
Example Problems
1. A soccer ball is kicked with a speed of 20 m/s at an angle of 30° above the horizontal.
Step 2: Time to reach max height using \( v = v_0 + at \):
At the top, vertical velocity \( v_y = 0 \)
\( 0 = 10 - 9.8t \Rightarrow t = \frac{10}{9.8} = 1.02\,\text{s} \)
Total time of flight is double that: \( T = 2t = 2.04\,\text{s} \)
Step 3: Max height using \( y = v_0 t + \frac{1}{2} a t^2 \):
\( y = 10(1.02) + \frac{1}{2}(-9.8)(1.02)^2 = 10.2 - 5.1 = 5.1\,\text{m} \)
Step 4: Range using \( x = v_{0x} T \):
\( R = 17.32 \cdot 2.04 = 35.36\,\text{m} \)
2. A cannon fires a projectile at 45° with an initial velocity of 28 m/s.
Step 2: Time to fall using \( y = y_0 + v_{0y} t + \frac{1}{2} a t^2 \):
\( 0 = 80 + 0 \cdot t + \frac{1}{2}(-9.8)t^2 \)
\( 0 = 80 - 4.9t^2 \Rightarrow t^2 = \frac{80}{4.9} = 16.33 \)
\( t = \sqrt{16.33} = 4.04\,\text{s} \)
Step 3: Range using \( x = v_{0x} t \):
\( R = 15 \cdot 4.04 = 60.6\,\text{m} \)
4. A ball is launched at 30° above the horizontal from the top of a 50 m hill with an initial velocity of 25 m/s. Find the maximum height of the ball above the ground.
Step 2: Time to reach max height from \( v = v_0 + at \):
At the peak, vertical velocity \( v_y = 0 \)
\( 0 = 12.5 - 9.8t \Rightarrow t = \frac{12.5}{9.8} = 1.28\,\text{s} \)
Step 3: Height gained above launch point using \( y = v_{0y} t + \frac{1}{2} a t^2 \):
\( \Delta y = 12.5(1.28) + \frac{1}{2}(-9.8)(1.28)^2 \)
\( \Delta y = 16 - 8 = 8\,\text{m} \)
Step 4: Total max height above ground:
\( y_{max} = 50 + 8 = 58\,\text{m} \)
5. A projectile's horizontal position is given by \( x(t) = 15t \) meters and vertical position by \( y(t) = 20t - 4.9t^2 \) meters. Find the velocity at \( t = 1 \) second.
Step 1: Find horizontal velocity by taking the derivative of x(t)
\( v_x(t) = \frac{dx}{dt} = \frac{d}{dt}(15t) = 15\,\text{m/s} \)
Step 2: Find vertical velocity by taking the derivative of y(t)
\( v_y(t) = \frac{dy}{dt} = \frac{d}{dt}(20t - 4.9t^2) = 20 - 9.8t\,\text{m/s} \)
Step 3: Evaluate at \( t = 1 \) second
\( v_x(1) = 15\,\text{m/s} \) (constant)
\( v_y(1) = 20 - 9.8(1) = 10.2\,\text{m/s} \)