AP Physics 1 · Lesson 3

Projectiles in two dimensions

Know Know it

Once a projectile leaves the launcher, gravity is the only force on it (ignoring air), so its acceleration is 9.8 m/s² straight down. That splits the motion into two independent parts. Horizontally the velocity is constant, Δx=vxt\Delta x = v_xt. Vertically it is free fall: with up positive, vy=v0y−gtv_y = v_{0y} - gt and Δy=v0yt−12gt2\Delta y = v_{0y}t - \tfrac{1}{2}gt^2.

Time is the link. Find the time from the vertical motion, then use it in the horizontal motion. For a launch at angle θ above the horizontal with speed v0v_0, the components are v0x=v0cos⁡θv_{0x} = v_0\cos\theta and v0y=v0sin⁡θv_{0y} = v_0\sin\theta.

A horizontal launch has v0y=0v_{0y} = 0, so the fall time depends only on the height: t=2h/gt = \sqrt{2h/g}.

Horizontal and vertical motion of a projectile
QuantityHorizontal (x)Vertical (y, up positive)
Acceleration0-9.8 m/s²
VelocityConstant: v0xv_{0x}vy=v0y−gtv_y = v_{0y} - gt
DisplacementΔx=v0xt\Delta x = v_{0x}tΔy=v0yt−12gt2\Delta y = v_{0y}t - \tfrac{1}{2}gt^2

Worked example Easy

A stone is thrown horizontally at 6.0 m/s from a bridge 7.2 m above a river. Neglecting air resistance, how far from the point directly below the throw does the stone hit the water?

  1. 5.1 m
  2. 7.3 mAnswer
  3. 10 m
  4. 43 m

How to solve it

  1. Vertical: the stone starts with v0y=0v_{0y} = 0 and falls 7.2 m, so t=2h/g=14.4/9.8=1.21t = \sqrt{2h/g} = \sqrt{14.4/9.8} = 1.21 s.
  2. Horizontal: a constant 6.0 m/s for 1.21 s.
  3. Δx=(6.0)(1.21)≈7.3\Delta x = (6.0)(1.21) \approx 7.3 m.

Why each choice is right or wrong

  • A. Incorrect. 5.1 m drops the 2 in t=2h/gt = \sqrt{2h/g}, which makes the fall time too short.
  • B. Correct. The fall time is 2(7.2)/9.8=1.21\sqrt{2(7.2)/9.8} = 1.21 s, and the stone moves sideways at a steady 6.0 m/s, so it hits (6.0)(1.21) ≈ 7.3 m out.
  • C. Incorrect. 10 m puts 4h, 28.8 m, under the square root in place of 2h, which makes the fall time too long.
  • D. Incorrect. 43 m multiplies the speed by the height, 6.0 times 7.2, which is not a distance the stone travels.

Trap The top of the path is not a stop

At the highest point only the vertical velocity is zero. The horizontal velocity is the same as at launch, so the speed at the top equals v0xv_{0x}, and the acceleration is still 9.8 m/s² downward.

Explain Explain it

Task verbs: calculate, justify

For a launch from level ground, the time up equals the time down, so the time of flight is 2v0y/g2v_{0y}/g and the range is v0xv_{0x} times that time. The maximum height comes from the vertical motion alone: hmax⁡=v0y2/(2g)h_{\max} = v_{0y}^2/(2g).

When you justify a projectile answer, say which direction of the motion you used and name the principle behind your step, not just the formula.

Worked example Hard

A fountain nozzle at ground level shoots water at 7.0 m/s at 60° above the horizontal. Neglecting air resistance, how far from the nozzle does the water land on the level ground?

  1. 2.2 m
  2. 4.3 mAnswer
  3. 5.0 m
  4. 8.7 m

How to solve it

  1. Split the launch velocity: v0x=3.5v_{0x} = 3.5 m/s and v0y=6.06v_{0y} = 6.06 m/s.
  2. Vertical: the rise takes 6.06/9.8 = 0.62 s, so the time of flight is 1.24 s.
  3. Horizontal: Δx=(3.5)(1.24)≈4.3\Delta x = (3.5)(1.24) \approx 4.3 m.

Why each choice is right or wrong

  • A. Incorrect. 2.2 m uses only the time to reach the top of the arc, which is half the time of flight.
  • B. Correct. v0x=7.0cos⁡60∘=3.5v_{0x} = 7.0\cos 60^\circ = 3.5 m/s and v0y=7.0sin⁡60∘=6.06v_{0y} = 7.0\sin 60^\circ = 6.06 m/s. The time of flight is 2(6.06)/9.8=1.242(6.06)/9.8 = 1.24 s, so the range is (3.5)(1.24) ≈ 4.3 m.
  • C. Incorrect. 5.0 m is v02/gv_0^2/g, the longest range possible at this speed, which needs a 45° launch, not 60°.
  • D. Incorrect. 8.7 m uses 2sin⁡60∘2\sin 60^\circ in place of sin⁡(2×60∘)\sin(2 \times 60^\circ), which doubles the range.

Justify Your turn

Two balls leave level ground at the same moment. Ball 1 has velocity components of 4.0 m/s horizontally and 9.8 m/s upward; ball 2 has 8.0 m/s horizontally and 9.8 m/s upward. Neglect air resistance and use g = 9.8 m/s². Which ball lands first? Justify your answer, then calculate how far from its launch point each ball lands.

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Compare with a model answer

A full answer does this

  • Says the two balls land at the same time
  • Justifies it: the time in the air is set by the vertical motion, which is the same for both balls
  • Gives ball 1’s landing distance as 8.0 m
  • Gives ball 2’s landing distance as 16 m

Model answer

They land at the same time. A projectile’s time in the air is set by its vertical motion alone, and both balls start with the same 9.8 m/s upward velocity and fall under the same gravity, so each is in the air for 2(9.8)/9.8=2.02(9.8)/9.8 = 2.0 s. The horizontal velocity decides only how far each one goes: ball 1 lands (4.0)(2.0) = 8.0 m away and ball 2 lands (8.0)(2.0) = 16 m away.

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