Projectile Motion worksheet

1. Range (horizontal distance): 208.11 m

1. Use the surface gravity of Mars

   $g$$=3.71\text{\hspace{0.17em}m/s²}$

2. Find the sideways part of the launch speed, which stays the same for the whole flight: vₓ = v₀ × cos θ

   $28\times \cos \left(50\text{\hspace{0.17em}°}\right)$$=17.998053\text{\hspace{0.17em}m/s}$

3. Find the upward part of the launch speed, which gravity works against: v_y = v₀ × sin θ

   $28\times \sin \left(50\text{\hspace{0.17em}°}\right)$$=21.449244\text{\hspace{0.17em}m/s}$

4. Find the time in the air, the moment the height returns to the landing level: t = (v_y + √(v_y² + 2 g h₀)) ÷ g

   $\frac{21.449244+\sqrt{{21.449244}^2+2\times 3.71\times 0}}{3.71}$$=11.562935\text{\hspace{0.17em}s}$

5. Find the highest point, reached when the upward speed falls to zero: H = h₀ + v_y² ÷ (2g)

   $0+\frac{{21.449244}^2}{2\times 3.71}$$=62.004055\text{\hspace{0.17em}m}$

6. Find the range: the sideways speed never changes, so multiply it by the time in the air, R = vₓ × t

   $17.998053\times 11.562935$$=208.110318\text{\hspace{0.17em}m}$

7. Find the up-and-down speed at landing, since gravity has been slowing then speeding it the whole time: v_y − g × t

   $21.449244-3.71\times 11.562935$$=\mathrm{-21.449244}\text{\hspace{0.17em}m/s}$

8. Combine the sideways and up-and-down speeds to get the total speed at impact: √(vₓ² + v_y²)

   $\sqrt{{17.998053}^2+{\left(\mathrm{-21.449244}\right)}^2}$$=28\text{\hspace{0.17em}m/s}$

9. Find how steeply it comes down, measured below the horizontal: arctan(|v_y| ÷ vₓ)

   $\arctan \left(\frac{21.449244}{17.998053}\right)$$=50\text{\hspace{0.17em}°}$

2. Range (horizontal distance): 22.61 m

1. Use the surface gravity of Earth (standard gravity)

   $g$$=9.80665\text{\hspace{0.17em}m/s²}$

2. Find the sideways part of the launch speed, which stays the same for the whole flight: vₓ = v₀ × cos θ

   $16\times \cos \left(30\text{\hspace{0.17em}°}\right)$$=13.856406\text{\hspace{0.17em}m/s}$

3. Find the upward part of the launch speed, which gravity works against: v_y = v₀ × sin θ

   $16\times \sin \left(30\text{\hspace{0.17em}°}\right)$$=8\text{\hspace{0.17em}m/s}$

4. Find the time in the air, the moment the height returns to the landing level: t = (v_y + √(v_y² + 2 g h₀)) ÷ g

   $\frac{8+\sqrt{8^2+2\times 9.80665\times 0}}{9.80665}$$=1.631546\text{\hspace{0.17em}s}$

5. Find the highest point, reached when the upward speed falls to zero: H = h₀ + v_y² ÷ (2g)

   $0+\frac{8^2}{2\times 9.80665}$$=3.263092\text{\hspace{0.17em}m}$

6. Find the range: the sideways speed never changes, so multiply it by the time in the air, R = vₓ × t

   $13.856406\times 1.631546$$=22.607364\text{\hspace{0.17em}m}$

7. Find the up-and-down speed at landing, since gravity has been slowing then speeding it the whole time: v_y − g × t

   $8-9.80665\times 1.631546$$=\mathrm{-8}\text{\hspace{0.17em}m/s}$

8. Combine the sideways and up-and-down speeds to get the total speed at impact: √(vₓ² + v_y²)

   $\sqrt{{13.856406}^2+{\left(\mathrm{-8}\right)}^2}$$=16\text{\hspace{0.17em}m/s}$

9. Find how steeply it comes down, measured below the horizontal: arctan(|v_y| ÷ vₓ)

   $\arctan \left(\frac{8}{13.856406}\right)$$=30\text{\hspace{0.17em}°}$

3. Range (horizontal distance): 134.07 yd

1. Use the surface gravity of Mars

   $g$$=3.71\text{\hspace{0.17em}m/s²}$

2. Find the sideways part of the launch speed, which stays the same for the whole flight: vₓ = v₀ × cos θ

   $22\times \cos \left(55\text{\hspace{0.17em}°}\right)$$=12.618682\text{\hspace{0.17em}m/s}$

3. Find the upward part of the launch speed, which gravity works against: v_y = v₀ × sin θ

   $22\times \sin \left(55\text{\hspace{0.17em}°}\right)$$=18.021345\text{\hspace{0.17em}m/s}$

4. Find the time in the air, the moment the height returns to the landing level: t = (v_y + √(v_y² + 2 g h₀)) ÷ g

   $\frac{18.021345+\sqrt{{18.021345}^2+2\times 3.71\times 0}}{3.71}$$=9.715011\text{\hspace{0.17em}s}$

5. Find the highest point, reached when the upward speed falls to zero: H = h₀ + v_y² ÷ (2g)

   $0+\frac{{18.021345}^2}{2\times 3.71}$$=43.76939\text{\hspace{0.17em}m}$

6. Find the range: the sideways speed never changes, so multiply it by the time in the air, R = vₓ × t

   $12.618682\times 9.715011$$=122.590628\text{\hspace{0.17em}m}$

7. Change the answer to yd, the unit you picked: divide by 0.9144

   $\frac{122.590628}{0.9144}$$=134.066741\text{\hspace{0.17em}yd}$

8. Find the up-and-down speed at landing, since gravity has been slowing then speeding it the whole time: v_y − g × t

   $18.021345-3.71\times 9.715011$$=\mathrm{-18.021345}\text{\hspace{0.17em}m/s}$

9. Combine the sideways and up-and-down speeds to get the total speed at impact: √(vₓ² + v_y²)

   $\sqrt{{12.618682}^2+{\left(\mathrm{-18.021345}\right)}^2}$$=22\text{\hspace{0.17em}m/s}$

10. Find how steeply it comes down, measured below the horizontal: arctan(|v_y| ÷ vₓ)

    $\arctan \left(\frac{18.021345}{12.618682}\right)$$=55\text{\hspace{0.17em}°}$

4. Range (horizontal distance): 28.11 yd

1. Use the surface gravity of Earth (standard gravity)

   $g$$=9.80665\text{\hspace{0.17em}m/s²}$

2. Find the sideways part of the launch speed, which stays the same for the whole flight: vₓ = v₀ × cos θ

   $16\times \cos \left(50\text{\hspace{0.17em}°}\right)$$=10.284602\text{\hspace{0.17em}m/s}$

3. Find the upward part of the launch speed, which gravity works against: v_y = v₀ × sin θ

   $16\times \sin \left(50\text{\hspace{0.17em}°}\right)$$=12.256711\text{\hspace{0.17em}m/s}$

4. Find the time in the air, the moment the height returns to the landing level: t = (v_y + √(v_y² + 2 g h₀)) ÷ g

   $\frac{12.256711+\sqrt{{12.256711}^2+2\times 9.80665\times 0}}{9.80665}$$=2.499673\text{\hspace{0.17em}s}$

5. Find the highest point, reached when the upward speed falls to zero: H = h₀ + v_y² ÷ (2g)

   $0+\frac{{12.256711}^2}{2\times 9.80665}$$=7.659444\text{\hspace{0.17em}m}$

6. Find the range: the sideways speed never changes, so multiply it by the time in the air, R = vₓ × t

   $10.284602\times 2.499673$$=25.708145\text{\hspace{0.17em}m}$

7. Change the answer to yd, the unit you picked: divide by 0.9144

   $\frac{25.708145}{0.9144}$$=28.11477\text{\hspace{0.17em}yd}$

8. Find the up-and-down speed at landing, since gravity has been slowing then speeding it the whole time: v_y − g × t

   $12.256711-9.80665\times 2.499673$$=\mathrm{-12.256711}\text{\hspace{0.17em}m/s}$

9. Combine the sideways and up-and-down speeds to get the total speed at impact: √(vₓ² + v_y²)

   $\sqrt{{10.284602}^2+{\left(\mathrm{-12.256711}\right)}^2}$$=16\text{\hspace{0.17em}m/s}$

10. Find how steeply it comes down, measured below the horizontal: arctan(|v_y| ÷ vₓ)

    $\arctan \left(\frac{12.256711}{10.284602}\right)$$=50\text{\hspace{0.17em}°}$

5. Range (horizontal distance): 0.06 km

1. Use the gravity value you entered

   $g$$=13\text{\hspace{0.17em}m/s²}$

2. Find the sideways part of the launch speed, which stays the same for the whole flight: vₓ = v₀ × cos θ

   $28\times \cos \left(38\text{\hspace{0.17em}°}\right)$$=22.064301\text{\hspace{0.17em}m/s}$

3. Find the upward part of the launch speed, which gravity works against: v_y = v₀ × sin θ

   $28\times \sin \left(38\text{\hspace{0.17em}°}\right)$$=17.238521\text{\hspace{0.17em}m/s}$

4. Find the time in the air, the moment the height returns to the landing level: t = (v_y + √(v_y² + 2 g h₀)) ÷ g

   $\frac{17.238521+\sqrt{{17.238521}^2+2\times 13\times 0}}{13}$$=2.65208\text{\hspace{0.17em}s}$

5. Find the highest point, reached when the upward speed falls to zero: H = h₀ + v_y² ÷ (2g)

   $0+\frac{{17.238521}^2}{2\times 13}$$=11.429485\text{\hspace{0.17em}m}$

6. Find the range: the sideways speed never changes, so multiply it by the time in the air, R = vₓ × t

   $22.064301\times 2.65208$$=58.516296\text{\hspace{0.17em}m}$

7. Change the answer to km, the unit you picked: divide by 1,000

   $\frac{58.516296}{\mathrm{1,000}}$$=0.058516\text{\hspace{0.17em}km}$

8. Find the up-and-down speed at landing, since gravity has been slowing then speeding it the whole time: v_y − g × t

   $17.238521-13\times 2.65208$$=\mathrm{-17.238521}\text{\hspace{0.17em}m/s}$

9. Combine the sideways and up-and-down speeds to get the total speed at impact: √(vₓ² + v_y²)

   $\sqrt{{22.064301}^2+{\left(\mathrm{-17.238521}\right)}^2}$$=28\text{\hspace{0.17em}m/s}$

10. Find how steeply it comes down, measured below the horizontal: arctan(|v_y| ÷ vₓ)

    $\arctan \left(\frac{17.238521}{22.064301}\right)$$=38\text{\hspace{0.17em}°}$

6. Range (horizontal distance): 22.04 m

1. Use the gravity value you entered

   $g$$=11.5\text{\hspace{0.17em}m/s²}$

2. Find the sideways part of the launch speed, which stays the same for the whole flight: vₓ = v₀ × cos θ

   $16\times \cos \left(49\text{\hspace{0.17em}°}\right)$$=10.496944\text{\hspace{0.17em}m/s}$

3. Find the upward part of the launch speed, which gravity works against: v_y = v₀ × sin θ

   $16\times \sin \left(49\text{\hspace{0.17em}°}\right)$$=12.075353\text{\hspace{0.17em}m/s}$

4. Find the time in the air, the moment the height returns to the landing level: t = (v_y + √(v_y² + 2 g h₀)) ÷ g

   $\frac{12.075353+\sqrt{{12.075353}^2+2\times 11.5\times 0}}{11.5}$$=2.100061\text{\hspace{0.17em}s}$

5. Find the highest point, reached when the upward speed falls to zero: H = h₀ + v_y² ÷ (2g)

   $0+\frac{{12.075353}^2}{2\times 11.5}$$=6.339746\text{\hspace{0.17em}m}$

6. Find the range: the sideways speed never changes, so multiply it by the time in the air, R = vₓ × t

   $10.496944\times 2.100061$$=22.044228\text{\hspace{0.17em}m}$

7. Find the up-and-down speed at landing, since gravity has been slowing then speeding it the whole time: v_y − g × t

   $12.075353-11.5\times 2.100061$$=\mathrm{-12.075353}\text{\hspace{0.17em}m/s}$

8. Combine the sideways and up-and-down speeds to get the total speed at impact: √(vₓ² + v_y²)

   $\sqrt{{10.496944}^2+{\left(\mathrm{-12.075353}\right)}^2}$$=16\text{\hspace{0.17em}m/s}$

9. Find how steeply it comes down, measured below the horizontal: arctan(|v_y| ÷ vₓ)

   $\arctan \left(\frac{12.075353}{10.496944}\right)$$=49\text{\hspace{0.17em}°}$

7. Range (horizontal distance): 57.46 yd

1. Use the surface gravity of Mars

   $g$$=3.71\text{\hspace{0.17em}m/s²}$

2. Find the sideways part of the launch speed, which stays the same for the whole flight: vₓ = v₀ × cos θ

   $14\times \cos \left(48\text{\hspace{0.17em}°}\right)$$=9.367828\text{\hspace{0.17em}m/s}$

3. Find the upward part of the launch speed, which gravity works against: v_y = v₀ × sin θ

   $14\times \sin \left(48\text{\hspace{0.17em}°}\right)$$=10.404028\text{\hspace{0.17em}m/s}$

4. Find the time in the air, the moment the height returns to the landing level: t = (v_y + √(v_y² + 2 g h₀)) ÷ g

   $\frac{10.404028+\sqrt{{10.404028}^2+2\times 3.71\times 0}}{3.71}$$=5.60864\text{\hspace{0.17em}s}$

5. Find the highest point, reached when the upward speed falls to zero: H = h₀ + v_y² ÷ (2g)

   $0+\frac{{10.404028}^2}{2\times 3.71}$$=14.588112\text{\hspace{0.17em}m}$

6. Find the range: the sideways speed never changes, so multiply it by the time in the air, R = vₓ × t

   $9.367828\times 5.60864$$=52.540779\text{\hspace{0.17em}m}$

7. Change the answer to yd, the unit you picked: divide by 0.9144

   $\frac{52.540779}{0.9144}$$=57.459295\text{\hspace{0.17em}yd}$

8. Find the up-and-down speed at landing, since gravity has been slowing then speeding it the whole time: v_y − g × t

   $10.404028-3.71\times 5.60864$$=\mathrm{-10.404028}\text{\hspace{0.17em}m/s}$

9. Combine the sideways and up-and-down speeds to get the total speed at impact: √(vₓ² + v_y²)

   $\sqrt{{9.367828}^2+{\left(\mathrm{-10.404028}\right)}^2}$$=14\text{\hspace{0.17em}m/s}$

10. Find how steeply it comes down, measured below the horizontal: arctan(|v_y| ÷ vₓ)

    $\arctan \left(\frac{10.404028}{9.367828}\right)$$=48\text{\hspace{0.17em}°}$

8. Range (horizontal distance): 0.33 km

1. Use the surface gravity of the Moon

   $g$$=1.62\text{\hspace{0.17em}m/s²}$

2. Find the sideways part of the launch speed, which stays the same for the whole flight: vₓ = v₀ × cos θ

   $23\times \cos \left(45\text{\hspace{0.17em}°}\right)$$=16.263456\text{\hspace{0.17em}m/s}$

3. Find the upward part of the launch speed, which gravity works against: v_y = v₀ × sin θ

   $23\times \sin \left(45\text{\hspace{0.17em}°}\right)$$=16.263456\text{\hspace{0.17em}m/s}$

4. Find the time in the air, the moment the height returns to the landing level: t = (v_y + √(v_y² + 2 g h₀)) ÷ g

   $\frac{16.263456+\sqrt{{16.263456}^2+2\times 1.62\times 0}}{1.62}$$=20.078341\text{\hspace{0.17em}s}$

5. Find the highest point, reached when the upward speed falls to zero: H = h₀ + v_y² ÷ (2g)

   $0+\frac{{16.263456}^2}{2\times 1.62}$$=81.635802\text{\hspace{0.17em}m}$

6. Find the range: the sideways speed never changes, so multiply it by the time in the air, R = vₓ × t

   $16.263456\times 20.078341$$=326.54321\text{\hspace{0.17em}m}$

7. Change the answer to km, the unit you picked: divide by 1,000

   $\frac{326.54321}{\mathrm{1,000}}$$=0.326543\text{\hspace{0.17em}km}$

8. Find the up-and-down speed at landing, since gravity has been slowing then speeding it the whole time: v_y − g × t

   $16.263456-1.62\times 20.078341$$=\mathrm{-16.263456}\text{\hspace{0.17em}m/s}$

9. Combine the sideways and up-and-down speeds to get the total speed at impact: √(vₓ² + v_y²)

   $\sqrt{{16.263456}^2+{\left(\mathrm{-16.263456}\right)}^2}$$=23\text{\hspace{0.17em}m/s}$

10. Find how steeply it comes down, measured below the horizontal: arctan(|v_y| ÷ vₓ)

    $\arctan \left(\frac{16.263456}{16.263456}\right)$$=45\text{\hspace{0.17em}°}$

9. Range (horizontal distance): 0.01 km

1. Use the gravity value you entered

   $g$$=13\text{\hspace{0.17em}m/s²}$

2. Find the sideways part of the launch speed, which stays the same for the whole flight: vₓ = v₀ × cos θ

   $13\times \cos \left(48\text{\hspace{0.17em}°}\right)$$=8.698698\text{\hspace{0.17em}m/s}$

3. Find the upward part of the launch speed, which gravity works against: v_y = v₀ × sin θ

   $13\times \sin \left(48\text{\hspace{0.17em}°}\right)$$=9.660883\text{\hspace{0.17em}m/s}$

4. Find the time in the air, the moment the height returns to the landing level: t = (v_y + √(v_y² + 2 g h₀)) ÷ g

   $\frac{9.660883+\sqrt{{9.660883}^2+2\times 13\times 0}}{13}$$=1.48629\text{\hspace{0.17em}s}$

5. Find the highest point, reached when the upward speed falls to zero: H = h₀ + v_y² ÷ (2g)

   $0+\frac{{9.660883}^2}{2\times 13}$$=3.589718\text{\hspace{0.17em}m}$

6. Find the range: the sideways speed never changes, so multiply it by the time in the air, R = vₓ × t

   $8.698698\times 1.48629$$=12.928785\text{\hspace{0.17em}m}$

7. Change the answer to km, the unit you picked: divide by 1,000

   $\frac{12.928785}{\mathrm{1,000}}$$=0.012929\text{\hspace{0.17em}km}$

8. Find the up-and-down speed at landing, since gravity has been slowing then speeding it the whole time: v_y − g × t

   $9.660883-13\times 1.48629$$=\mathrm{-9.660883}\text{\hspace{0.17em}m/s}$

9. Combine the sideways and up-and-down speeds to get the total speed at impact: √(vₓ² + v_y²)

   $\sqrt{{8.698698}^2+{\left(\mathrm{-9.660883}\right)}^2}$$=13\text{\hspace{0.17em}m/s}$

10. Find how steeply it comes down, measured below the horizontal: arctan(|v_y| ÷ vₓ)

    $\arctan \left(\frac{9.660883}{8.698698}\right)$$=48\text{\hspace{0.17em}°}$

10. Range (horizontal distance): 0.24 km

1. Use the surface gravity of the Moon

   $g$$=1.62\text{\hspace{0.17em}m/s²}$

2. Find the sideways part of the launch speed, which stays the same for the whole flight: vₓ = v₀ × cos θ

   $20\times \cos \left(40\text{\hspace{0.17em}°}\right)$$=15.320889\text{\hspace{0.17em}m/s}$

3. Find the upward part of the launch speed, which gravity works against: v_y = v₀ × sin θ

   $20\times \sin \left(40\text{\hspace{0.17em}°}\right)$$=12.855752\text{\hspace{0.17em}m/s}$

4. Find the time in the air, the moment the height returns to the landing level: t = (v_y + √(v_y² + 2 g h₀)) ÷ g

   $\frac{12.855752+\sqrt{{12.855752}^2+2\times 1.62\times 0}}{1.62}$$=15.871299\text{\hspace{0.17em}s}$

5. Find the highest point, reached when the upward speed falls to zero: H = h₀ + v_y² ÷ (2g)

   $0+\frac{{12.855752}^2}{2\times 1.62}$$=51.009372\text{\hspace{0.17em}m}$

6. Find the range: the sideways speed never changes, so multiply it by the time in the air, R = vₓ × t

   $15.320889\times 15.871299$$=243.162408\text{\hspace{0.17em}m}$

7. Change the answer to km, the unit you picked: divide by 1,000

   $\frac{243.162408}{\mathrm{1,000}}$$=0.243162\text{\hspace{0.17em}km}$

8. Find the up-and-down speed at landing, since gravity has been slowing then speeding it the whole time: v_y − g × t

   $12.855752-1.62\times 15.871299$$=\mathrm{-12.855752}\text{\hspace{0.17em}m/s}$

9. Combine the sideways and up-and-down speeds to get the total speed at impact: √(vₓ² + v_y²)

   $\sqrt{{15.320889}^2+{\left(\mathrm{-12.855752}\right)}^2}$$=20\text{\hspace{0.17em}m/s}$

10. Find how steeply it comes down, measured below the horizontal: arctan(|v_y| ÷ vₓ)

    $\arctan \left(\frac{12.855752}{15.320889}\right)$$=40\text{\hspace{0.17em}°}$