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Projectile Motion

Dynamics · FE Reference Handbook section

Dynamics
6 formulas
10 exam-style examples
~57 min
All Dynamics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

  • The equations for common projectile motion may be obtained from the constant acceleration equations as

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1
Projectile range — solve for range — Projectile Motion

A dynamics problem uses Projectile range. Given launch speed (v0) = 41.0000 m/s; launch angle (theta) = 61.0000 deg, determine the range (R) in m.

Given

  • launchspeed(v0)=41.0000m/slaunch speed (v_{0}) = 41.0000 m/s
  • launchangle(theta)=61.0000deglaunch angle (theta) = 61.0000 deg

Find

range (R), in m

Start with the thinking

  • The governing relation printed in this handbook section is Projectile range.
  • Everything except R is given, so isolate R symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=v02sin⁡(2θ)/gR = v_0^2 \sin(2\theta) / g
  2. Step 2 — Rearrange the relation so that R stands alone on the left-hand side.

  3. Step 3

    Listthegivens:launchspeed(v0)=41.0000m/s,launchangle(theta)=61.0000degList the givens: launch speed (v_{0}) = 41.0000 m/s, launch angle (theta) = 61.0000 deg
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    R=145.3 mR = 145.3\ \text{m}
  6. Step 6 — Check: returning R = 145.3 m to

    R=v02sin⁡(2θ)/gR = v_0^2 \sin(2\theta) / g

    reproduces the given quantities, and both sides carry the same units.

Answer:
R=145.3 mR = 145.3\ \text{m}

Why the other options are there

  • 290.6 — kept a factor of two that cancels in the correct rearrangement.
  • 72.6590 — dropped that same factor in the other direction.
  • 159.8 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Projectile Motion

Example 2
Projectile range — solve for launch speed — Projectile Motion (2)

A dynamics problem uses Projectile range. Given launch angle (theta) = 67.0000 deg; range (R) = 134.5 m, determine the launch speed (v0) in m/s.

Given

  • launchangle(theta)=67.0000deglaunch angle (theta) = 67.0000 deg
  • range(R)=134.5mrange (R) = 134.5 m

Find

launch speed (v0), in m/s

Start with the thinking

  • The governing relation printed in this handbook section is Projectile range.
  • Everything except v0 is given, so isolate v0 symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=v02sin⁡(2θ)/gR = v_0^2 \sin(2\theta) / g
  2. Step 2 — Rearrange the relation so that v0 stands alone on the left-hand side.

  3. Step 3

    Listthegivens:launchangle(theta)=67.0000deg,range(R)=134.5mList the givens: launch angle (theta) = 67.0000 deg, range (R) = 134.5 m
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    v0=42.8281 m/sv_{0} = 42.8281\ \text{m/s}
  6. Step 6 — Check: returning v0 = 42.8281 m/s to

    R=v02sin⁡(2θ)/gR = v_0^2 \sin(2\theta) / g

    reproduces the given quantities, and both sides carry the same units.

Answer:
v0=42.8281 m/sv_{0} = 42.8281\ \text{m/s}

Why the other options are there

  • 85.6562 — kept a factor of two that cancels in the correct rearrangement.
  • 21.4140 — dropped that same factor in the other direction.
  • 47.1109 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Projectile Motion

Example 3
Projectile range — solve for range (case 2) — Projectile Motion (3)

A dynamics problem uses Projectile range. Given launch speed (v0) = 16.5000 m/s; launch angle (theta) = 65.0000 deg, determine the range (R) in m.

Given

  • launchspeed(v0)=16.5000m/slaunch speed (v_{0}) = 16.5000 m/s
  • launchangle(theta)=65.0000deglaunch angle (theta) = 65.0000 deg

Find

range (R), in m

Start with the thinking

  • The governing relation printed in this handbook section is Projectile range.
  • Everything except R is given, so isolate R symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=v02sin⁡(2θ)/gR = v_0^2 \sin(2\theta) / g
  2. Step 2 — Rearrange the relation so that R stands alone on the left-hand side.

  3. Step 3

    Listthegivens:launchspeed(v0)=16.5000m/s,launchangle(theta)=65.0000degList the givens: launch speed (v_{0}) = 16.5000 m/s, launch angle (theta) = 65.0000 deg
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    R=21.2595 mR = 21.2595\ \text{m}
  6. Step 6 — Check: returning R = 21.2595 m to

    R=v02sin⁡(2θ)/gR = v_0^2 \sin(2\theta) / g

    reproduces the given quantities, and both sides carry the same units.

Answer:
R=21.2595 mR = 21.2595\ \text{m}

Why the other options are there

  • 42.5190 — kept a factor of two that cancels in the correct rearrangement.
  • 10.6297 — dropped that same factor in the other direction.
  • 23.3854 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Projectile Motion

Example 4
Projectile range — solve for launch speed (case 2) — Projectile Motion (4)

A dynamics problem uses Projectile range. Given launch angle (theta) = 27.0000 deg; range (R) = 18.8000 m, determine the launch speed (v0) in m/s.

Given

  • launchangle(theta)=27.0000deglaunch angle (theta) = 27.0000 deg
  • range(R)=18.8000mrange (R) = 18.8000 m

Find

launch speed (v0), in m/s

Start with the thinking

  • The governing relation printed in this handbook section is Projectile range.
  • Everything except v0 is given, so isolate v0 symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=v02sin⁡(2θ)/gR = v_0^2 \sin(2\theta) / g
  2. Step 2 — Rearrange the relation so that v0 stands alone on the left-hand side.

  3. Step 3

    Listthegivens:launchangle(theta)=27.0000deg,range(R)=18.8000mList the givens: launch angle (theta) = 27.0000 deg, range (R) = 18.8000 m
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    v0=15.0985 m/sv_{0} = 15.0985\ \text{m/s}
  6. Step 6 — Check: returning v0 = 15.0985 m/s to

    R=v02sin⁡(2θ)/gR = v_0^2 \sin(2\theta) / g

    reproduces the given quantities, and both sides carry the same units.

Answer:
v0=15.0985 m/sv_{0} = 15.0985\ \text{m/s}

Why the other options are there

  • 30.1971 — kept a factor of two that cancels in the correct rearrangement.
  • 7.5493 — dropped that same factor in the other direction.
  • 16.6084 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Projectile Motion

Example 5
Projectile range — solve for range (case 3) — Projectile Motion (5)

A dynamics problem uses Projectile range. Given launch speed (v0) = 46.0000 m/s; launch angle (theta) = 11.0000 deg, determine the range (R) in m.

Given

  • launchspeed(v0)=46.0000m/slaunch speed (v_{0}) = 46.0000 m/s
  • launchangle(theta)=11.0000deglaunch angle (theta) = 11.0000 deg

Find

range (R), in m

Start with the thinking

  • The governing relation printed in this handbook section is Projectile range.
  • Everything except R is given, so isolate R symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=v02sin⁡(2θ)/gR = v_0^2 \sin(2\theta) / g
  2. Step 2 — Rearrange the relation so that R stands alone on the left-hand side.

  3. Step 3

    Listthegivens:launchspeed(v0)=46.0000m/s,launchangle(theta)=11.0000degList the givens: launch speed (v_{0}) = 46.0000 m/s, launch angle (theta) = 11.0000 deg
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    R=80.8020 mR = 80.8020\ \text{m}
  6. Step 6 — Check: returning R = 80.8020 m to

    R=v02sin⁡(2θ)/gR = v_0^2 \sin(2\theta) / g

    reproduces the given quantities, and both sides carry the same units.

Answer:
R=80.8020 mR = 80.8020\ \text{m}

Why the other options are there

  • 161.6 — kept a factor of two that cancels in the correct rearrangement.
  • 40.4010 — dropped that same factor in the other direction.
  • 88.8822 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Projectile Motion

Example 6
Projectile range — solve for launch speed (case 3) — Projectile Motion (6)

A dynamics problem uses Projectile range. Given launch angle (theta) = 60.0000 deg; range (R) = 128.4 m, determine the launch speed (v0) in m/s.

Given

  • launchangle(theta)=60.0000deglaunch angle (theta) = 60.0000 deg
  • range(R)=128.4mrange (R) = 128.4 m

Find

launch speed (v0), in m/s

Start with the thinking

  • The governing relation printed in this handbook section is Projectile range.
  • Everything except v0 is given, so isolate v0 symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=v02sin⁡(2θ)/gR = v_0^2 \sin(2\theta) / g
  2. Step 2 — Rearrange the relation so that v0 stands alone on the left-hand side.

  3. Step 3

    Listthegivens:launchangle(theta)=60.0000deg,range(R)=128.4mList the givens: launch angle (theta) = 60.0000 deg, range (R) = 128.4 m
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    v0=38.1375 m/sv_{0} = 38.1375\ \text{m/s}
  6. Step 6 — Check: returning v0 = 38.1375 m/s to

    R=v02sin⁡(2θ)/gR = v_0^2 \sin(2\theta) / g

    reproduces the given quantities, and both sides carry the same units.

Answer:
v0=38.1375 m/sv_{0} = 38.1375\ \text{m/s}

Why the other options are there

  • 76.2749 — kept a factor of two that cancels in the correct rearrangement.
  • 19.0687 — dropped that same factor in the other direction.
  • 41.9512 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Projectile Motion

Example 7
Projectile range — solve for range (case 4) — Projectile Motion (7)

A dynamics problem uses Projectile range. Given launch speed (v0) = 34.5000 m/s; launch angle (theta) = 16.0000 deg, determine the range (R) in m.

Given

  • launchspeed(v0)=34.5000m/slaunch speed (v_{0}) = 34.5000 m/s
  • launchangle(theta)=16.0000deglaunch angle (theta) = 16.0000 deg

Find

range (R), in m

Start with the thinking

  • The governing relation printed in this handbook section is Projectile range.
  • Everything except R is given, so isolate R symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=v02sin⁡(2θ)/gR = v_0^2 \sin(2\theta) / g
  2. Step 2 — Rearrange the relation so that R stands alone on the left-hand side.

  3. Step 3

    Listthegivens:launchspeed(v0)=34.5000m/s,launchangle(theta)=16.0000degList the givens: launch speed (v_{0}) = 34.5000 m/s, launch angle (theta) = 16.0000 deg
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    R=64.2953 mR = 64.2953\ \text{m}
  6. Step 6 — Check: returning R = 64.2953 m to

    R=v02sin⁡(2θ)/gR = v_0^2 \sin(2\theta) / g

    reproduces the given quantities, and both sides carry the same units.

Answer:
R=64.2953 mR = 64.2953\ \text{m}

Why the other options are there

  • 128.6 — kept a factor of two that cancels in the correct rearrangement.
  • 32.1476 — dropped that same factor in the other direction.
  • 70.7248 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Projectile Motion

Example 8
Projectile range — solve for launch speed (case 4) — Projectile Motion (8)

A dynamics problem uses Projectile range. Given launch angle (theta) = 44.0000 deg; range (R) = 268.9 m, determine the launch speed (v0) in m/s.

Given

  • launchangle(theta)=44.0000deglaunch angle (theta) = 44.0000 deg
  • range(R)=268.9mrange (R) = 268.9 m

Find

launch speed (v0), in m/s

Start with the thinking

  • The governing relation printed in this handbook section is Projectile range.
  • Everything except v0 is given, so isolate v0 symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=v02sin⁡(2θ)/gR = v_0^2 \sin(2\theta) / g
  2. Step 2 — Rearrange the relation so that v0 stands alone on the left-hand side.

  3. Step 3

    Listthegivens:launchangle(theta)=44.0000deg,range(R)=268.9mList the givens: launch angle (theta) = 44.0000 deg, range (R) = 268.9 m
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    v0=51.3762 m/sv_{0} = 51.3762\ \text{m/s}
  6. Step 6 — Check: returning v0 = 51.3762 m/s to

    R=v02sin⁡(2θ)/gR = v_0^2 \sin(2\theta) / g

    reproduces the given quantities, and both sides carry the same units.

Answer:
v0=51.3762 m/sv_{0} = 51.3762\ \text{m/s}

Why the other options are there

  • 102.8 — kept a factor of two that cancels in the correct rearrangement.
  • 25.6881 — dropped that same factor in the other direction.
  • 56.5139 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Projectile Motion

Example 9
Projectile range — solve for range (case 5) — Projectile Motion (9)

A dynamics problem uses Projectile range. Given launch speed (v0) = 29.0000 m/s; launch angle (theta) = 65.0000 deg, determine the range (R) in m.

Given

  • launchspeed(v0)=29.0000m/slaunch speed (v_{0}) = 29.0000 m/s
  • launchangle(theta)=65.0000deglaunch angle (theta) = 65.0000 deg

Find

range (R), in m

Start with the thinking

  • The governing relation printed in this handbook section is Projectile range.
  • Everything except R is given, so isolate R symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=v02sin⁡(2θ)/gR = v_0^2 \sin(2\theta) / g
  2. Step 2 — Rearrange the relation so that R stands alone on the left-hand side.

  3. Step 3

    Listthegivens:launchspeed(v0)=29.0000m/s,launchangle(theta)=65.0000degList the givens: launch speed (v_{0}) = 29.0000 m/s, launch angle (theta) = 65.0000 deg
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    R=65.6721 mR = 65.6721\ \text{m}
  6. Step 6 — Check: returning R = 65.6721 m to

    R=v02sin⁡(2θ)/gR = v_0^2 \sin(2\theta) / g

    reproduces the given quantities, and both sides carry the same units.

Answer:
R=65.6721 mR = 65.6721\ \text{m}

Why the other options are there

  • 131.3 — kept a factor of two that cancels in the correct rearrangement.
  • 32.8361 — dropped that same factor in the other direction.
  • 72.2393 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Projectile Motion

Example 10
Projectile range — solve for launch speed (case 5) — Projectile Motion (10)

A dynamics problem uses Projectile range. Given launch angle (theta) = 19.0000 deg; range (R) = 253.2 m, determine the launch speed (v0) in m/s.

Given

  • launchangle(theta)=19.0000deglaunch angle (theta) = 19.0000 deg
  • range(R)=253.2mrange (R) = 253.2 m

Find

launch speed (v0), in m/s

Start with the thinking

  • The governing relation printed in this handbook section is Projectile range.
  • Everything except v0 is given, so isolate v0 symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    R=v02sin⁡(2θ)/gR = v_0^2 \sin(2\theta) / g
  2. Step 2 — Rearrange the relation so that v0 stands alone on the left-hand side.

  3. Step 3

    Listthegivens:launchangle(theta)=19.0000deg,range(R)=253.2mList the givens: launch angle (theta) = 19.0000 deg, range (R) = 253.2 m
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    v0=63.5178 m/sv_{0} = 63.5178\ \text{m/s}
  6. Step 6 — Check: returning v0 = 63.5178 m/s to

    R=v02sin⁡(2θ)/gR = v_0^2 \sin(2\theta) / g

    reproduces the given quantities, and both sides carry the same units.

Answer:
v0=63.5178 m/sv_{0} = 63.5178\ \text{m/s}

Why the other options are there

  • 127.0 — kept a factor of two that cancels in the correct rearrangement.
  • 31.7589 — dropped that same factor in the other direction.
  • 69.8696 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Projectile Motion

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