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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.