Difficulty: Intermediate | Prerequisites: 1D kinematics, trigonometry (sine, cosine, tangent)
Projectile motion is the first place in the course where you work in two dimensions. The core idea is simple: horizontal and vertical motions are independent of each other. Gravity only acts vertically, so the horizontal component of velocity stays constant throughout the flight. If you can solve two separate 1D kinematics problems (one for x, one for y) and connect them through time, you can solve any projectile problem. This topic draws heavily on trigonometric decomposition of vectors.
A projectile is any object moving under gravity alone after launch. Split the motion into horizontal (constant velocity) and vertical (constant acceleration of -g). Time links the two. For level-ground problems, you can use the range and time-of-flight formulas directly. For problems where the launch and landing heights differ, go back to the component equations.
Projectile
Any object that moves through the air under the influence of gravity alone (no engine, no air resistance). Once launched, the only acceleration is g downward.
Launch angle (theta)
The angle between the initial velocity vector and the horizontal. A higher angle gives more vertical velocity but less horizontal velocity for the same speed.
Horizontal component of velocity (v0x)
v0x = v0 cos(theta). This stays constant throughout the flight because there is no horizontal acceleration.
Vertical component of velocity (v0y)
v0y = v0 sin(theta). This changes over time due to gravity: vy = v0y - gt.
Time of flight
The total time the projectile is in the air. For level ground (launch height = landing height), T = 2 v0 sin(theta) / g.
Range
The horizontal distance from launch to landing. For level ground, R = v0^2 sin(2 theta) / g.
Trajectory
The path followed by the projectile. In the absence of air resistance, it is a parabola.
Every projectile problem starts by breaking the launch velocity into components: v0x = v0 cos(theta), v0y = v0 sin(theta).
Horizontal: x = v0x * t (constant velocity, no acceleration).
Vertical: y = v0y * t - 0.5 g t^2, and vy = v0y - g t.
Time is the variable that connects horizontal and vertical motion. Solve for t in whichever equation gives you enough information, then use it in the other.
When launch and landing are at the same height, the flight is symmetric: time up = time down.
Time of flight: T = 2 v0 sin(theta) / g = 2(46.5) sin(37.5) / 9.80 = 2(46.5)(0.6088) / 9.80 = 56.62 / 9.80 = 5.78 s. Answer Q8: C) 5.77 s.
Range: R = v0x * T = v0 cos(theta) * T = 46.5 cos(37.5) * 5.78 = 46.5(0.7934)(5.78) = 213.2 m. Answer Q9: B) 213 m.
Alternatively, R = v0^2 sin(2 theta) / g = (46.5)^2 sin(75) / 9.80 = 2162.25(0.9659) / 9.80 = 213.1 m.
Three projectiles launched with the same speed but different angles: A has the steepest angle, C has the shallowest.
Horizontal component = v0 cos(theta). A larger angle means a smaller cosine, so a smaller horizontal component.
From the diagram, C has the smallest launch angle (most horizontal), A has the largest (most vertical).
Ranking in order of decreasing v0x: C > B > A. Answer: C) C > B > A.
The archerfish squirts water at 2.30 m/s at 19.5 degrees above horizontal. The water hits a beetle on a leaf at height h, where it is moving horizontally (meaning vy = 0 at the beetle).
Q13: Initial velocity components.
v0x = 2.30 cos(19.5) = 2.30(0.9426) = 2.17 m/s.
v0y = 2.30 sin(19.5) = 2.30(0.3338) = 0.768 m/s.
Q14: Reaction time (time for water to reach the beetle).
At the beetle, vy = 0. Use vy = v0y - gt: 0 = 0.768 - 9.80 t, so t = 0.768 / 9.80 = 0.0784 s.
The beetle has about 0.078 s to react.
Q15: Height h of the beetle.
h = v0y t - 0.5 g t^2 = 0.768(0.0784) - 0.5(9.80)(0.0784)^2 = 0.0602 - 0.0301 = 0.0301 m, about 3.0 cm.
Q16: Horizontal distance d.
d = v0x * t = 2.17(0.0784) = 0.170 m, about 17.0 cm.
The change in velocity for an interval is delta v = v_f - v_i. To find this graphically, reverse the direction of v_i and add it tip-to-tail to v_f.
Average acceleration = delta v / delta t. A larger delta v over a shorter time interval gives a larger average acceleration.
For ranking, compute the magnitude of delta v for each interval, then divide by the interval's duration. The interval with the largest ratio has the largest average acceleration magnitude.
v_{0x} = v_0 \cos\theta \qquad v_{0y} = v_0 \sin\thetax = v_{0x} \, t \qquad y = v_{0y} \, t - \tfrac{1}{2}g t^2v_y = v_{0y} - g t \qquad v_y^2 = v_{0y}^2 - 2 g yLevel-ground special cases:
T = \frac{2 v_0 \sin\theta}{g} \qquad R = \frac{v_0^2 \sin(2\theta)}{g}Projectile motion governs everything from basketball shots to water fountains to ballistic trajectories. Engineers use these equations to design irrigation sprinklers (maximising range at 45 degrees), artillery tables, and sports equipment. The archerfish problem is a real biological example: these fish have evolved to aim precisely by compensating for gravity's effect on the water stream.
Students often think horizontal velocity changes during flight. It does not (in the absence of air resistance). Only the vertical component changes.
At the peak of the trajectory, the velocity is not zero. The vertical component is zero, but the horizontal component is still v0 cos(theta). The speed at the peak equals v0 cos(theta).
Students sometimes use the range formula for problems where launch and landing heights are different. The range formula R = v0^2 sin(2 theta) / g only works when the projectile lands at the same height it was launched from.
"Moving horizontally" means vy = 0 at that instant. It does not mean the object has stopped.
Expect at least one full multi-part projectile problem on the exam (like the archerfish or golfer problems). Practise the full workflow: decompose, solve for time, then find range or height.
Ranking questions (like Q10) test whether you understand how angle affects the components. Remember: larger angle means more v0y, less v0x.
The condition "moving horizontally" (vy = 0) is a common exam trigger. It tells you the projectile is at its peak.
True or false: At the top of a projectile's arc, its acceleration is zero. (False. Acceleration is -g = -9.80 m/s^2 throughout the flight.)
A ball is launched at 30 degrees. Its horizontal velocity component is v0 cos(30) = ______. (0.866 v0)
Fill in the blank: For level-ground flight, the angle that maximises range is ______ degrees. (45)
True or false: A projectile launched at 60 degrees has the same range as one launched at 30 degrees (same speed). (True. sin(2 x 60) = sin(120) = sin(60) = sin(2 x 30).)
If a projectile is "moving horizontally" at some point, what is vy at that point? (0)
Q: A golfer hits a ball at 46.5 m/s at 37.5 degrees on level ground. How long is the ball in flight?
A: T = 2 v0 sin(theta) / g = 2(46.5)(sin 37.5) / 9.80 = 2(46.5)(0.6088) / 9.80 = 5.77 s.
Q: For the same golfer, how far does the ball land from the tee?
A: R = v0 cos(theta) x T = 46.5(0.7934)(5.77) = 213 m. Or use R = v0^2 sin(2 theta) / g.
Q: Three projectiles are launched at the same speed. A is at 75 degrees, B at 45 degrees, C at 30 degrees. Rank them by decreasing horizontal velocity component.
A: C > B > A. Smaller angle means larger cos(theta), so larger v0x.
Q: An archerfish squirts water at 2.30 m/s at 19.5 degrees. The water arrives at a beetle moving horizontally. Find the height of the beetle.
A: v0y = 2.30 sin(19.5) = 0.768 m/s. Time to reach vy = 0: t = 0.768/9.80 = 0.0784 s. Height: h = v0y t - 0.5 g t^2 = 0.768(0.0784) - 0.5(9.80)(0.0784)^2 = 0.030 m = 3.0 cm.
Q: What is the horizontal distance between the fish and the beetle at launch?
A: d = v0x x t = 2.30 cos(19.5) x 0.0784 = 2.17 x 0.0784 = 0.170 m = 17.0 cm.
Projectile motion is 1D kinematics applied twice (once horizontally, once vertically), so everything from the previous set of notes carries over. When you add forces in the next unit, you will see that the reason horizontal velocity is constant is that there is no horizontal force (Newton's first law). Energy methods will later give you a shortcut: the speed at any height can be found from conservation of energy without decomposing into components.
projectile motion, 2D kinematics, launch angle, range formula, time of flight, horizontal component, vertical component, trajectory, parabola, archerfish, symmetric flight, complementary angles, delta v, change in velocity, average acceleration, velocity vector subtraction, PHYSICS 114, UW general physics