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Projectile motion describes the curved path of an object launched into the air and moving under gravity. This cheat sheet helps students separate motion into horizontal and vertical parts so problems become easier to solve. It is useful for finding time, distance, height, velocity, and launch angle in common physics problems.

Students need these equations because projectile motion combines vectors, acceleration, and kinematics in one topic.

The key idea is that horizontal velocity stays constant when air resistance is ignored, while vertical velocity changes because of gravity. Initial velocity is split into components using v0x=v0cosθv_{0x} = v_0 \cos\theta and v0y=v0sinθv_{0y} = v_0 \sin\theta. Vertical motion uses ay=ga_y = -g, where g=9.8m/s2g = 9.8\,\text{m/s}^2 near Earth.

For level launches and landings, useful shortcuts include T=2v0sinθgT = \frac{2v_0\sin\theta}{g}, R=v02sin2θgR = \frac{v_0^2\sin 2\theta}{g}, and H=v02sin2θ2gH = \frac{v_0^2\sin^2\theta}{2g}.

Key Facts

  • Horizontal and vertical motion are independent, so solve xx motion and yy motion separately using the same time tt.
  • The horizontal velocity component is v0x=v0cosθv_{0x} = v_0\cos\theta, and the vertical velocity component is v0y=v0sinθv_{0y} = v_0\sin\theta.
  • With no air resistance, horizontal acceleration is ax=0a_x = 0, so horizontal position is x=x0+v0xtx = x_0 + v_{0x}t.
  • Vertical acceleration near Earth is ay=g=9.8m/s2a_y = -g = -9.8\,\text{m/s}^2 when upward is chosen as positive.
  • Vertical position is modeled by y=y0+v0yt+12ayt2y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2.
  • Vertical velocity is modeled by vy=v0y+aytv_y = v_{0y} + a_yt, and at maximum height vy=0v_y = 0.
  • For a projectile that lands at the same height it was launched, total flight time is T=2v0sinθgT = \frac{2v_0\sin\theta}{g}.
  • For a projectile that lands at the same height it was launched, range is R=v02sin2θgR = \frac{v_0^2\sin 2\theta}{g} and maximum height is H=v02sin2θ2gH = \frac{v_0^2\sin^2\theta}{2g}.

Vocabulary

Projectile
A projectile is an object that moves through the air after being launched and is acted on mainly by gravity.
Trajectory
A trajectory is the curved path followed by a projectile during its motion.
Launch Angle
The launch angle is the angle θ\theta between the initial velocity vector and the horizontal direction.
Initial Velocity Components
Initial velocity components are the horizontal and vertical parts of the launch velocity, given by v0x=v0cosθv_{0x} = v_0\cos\theta and v0y=v0sinθv_{0y} = v_0\sin\theta.
Range
Range is the horizontal distance a projectile travels, often represented by RR.
Maximum Height
Maximum height is the highest vertical position reached by a projectile, where its vertical velocity is vy=0v_y = 0.

Common Mistakes to Avoid

  • Using the full launch speed as both components is wrong because v0v_0 must be split into v0x=v0cosθv_{0x} = v_0\cos\theta and v0y=v0sinθv_{0y} = v_0\sin\theta.
  • Putting gg in the horizontal equation is wrong because gravity acts vertically, so ax=0a_x = 0 when air resistance is ignored.
  • Forgetting the sign of acceleration is wrong because if upward is positive, vertical acceleration must be ay=9.8m/s2a_y = -9.8\,\text{m/s}^2.
  • Using the range formula for unequal launch and landing heights is wrong because R=v02sin2θgR = \frac{v_0^2\sin 2\theta}{g} only applies when the projectile lands at its launch height.
  • Assuming velocity is zero at the top is wrong because only the vertical velocity is zero, so vy=0v_y = 0 but vxv_x is still constant.

Practice Questions

  1. 1 A ball is launched at 20m/s20\,\text{m/s} at an angle of 3030^\circ above the horizontal. Find v0xv_{0x} and v0yv_{0y}.
  2. 2 A projectile is launched horizontally from a cliff with v0x=12m/sv_{0x} = 12\,\text{m/s} and stays in the air for 3.0s3.0\,\text{s}. How far horizontally does it travel?
  3. 3 A projectile is launched from level ground at 25m/s25\,\text{m/s} and 4040^\circ. Using g=9.8m/s2g = 9.8\,\text{m/s}^2, find its approximate time of flight.
  4. 4 Explain why two projectiles launched with the same speed at angles 3030^\circ and 6060^\circ can have the same range when they land at the same height.

Understanding Projectile Motion Equations

The most important skill in projectile problems is choosing a coordinate system before inserting numbers. Many students choose right as positive for horizontal motion and up as positive for vertical motion. That choice is not mandatory, but it must be used consistently.

A downward velocity then has a negative value, even when the object is still moving quickly. The acceleration due to gravity remains negative throughout the trip. It does not become zero at the top of the path.

Only the vertical velocity becomes zero for one instant there. This distinction prevents a common mistake. An object at its highest point still has horizontal motion unless it was thrown straight up.

The same clock controls both directions. Start timing at launch and use one time value for every part of the motion. This is especially useful when a ball is launched from a balcony, a cliff, or a table.

In these cases, the landing height differs from the launch height, so the familiar level-ground shortcuts do not apply. Use the vertical position relationship to find the physically meaningful time when the object reaches the ground. There can be two mathematical time answers when the chosen ground level is reached before and after launch in a model.

A negative time describes an event before launch, so it must be rejected. Once the positive landing time is known, use horizontal motion to find where the object lands.

Graphs show the ideas in another useful way. A horizontal position versus time graph is a straight line when horizontal speed is constant. Its slope gives the horizontal velocity.

A vertical velocity versus time graph is a straight line that slopes downward. Its slope is the vertical acceleration. A vertical position versus time graph is curved, with a peak at the maximum height.

Reading these graphs helps students check whether an answer makes physical sense. For example, equal time intervals on the way up and down change vertical velocity by equal amounts when air resistance is ignored. The vertical speed at a given height matches on both sides of the path, but the direction is opposite.

Launch angle produces effects that can seem surprising. For launches and landings at the same height, two angles that add to ninety degrees give the same horizontal range when the launch speed is unchanged. One path is high and steep, while the other is low and fast.

A forty five degree launch gives the greatest range only under the ideal conditions of level ground and no air resistance. In real sports, the best angle depends on release height, landing height, spin, wind, and drag. A basketball shot, a kicked football, and water from a hose all depart from the ideal model.

The model still gives a strong first estimate. Check units at every stage.

Speeds use meters per second, acceleration uses meters per second per second, time uses seconds, and distance uses meters. Unit checks often reveal a wrong formula or an incorrect calculator entry.