Projectiles | Jamb(UTME)

Projectile Motion Basics

1. Projectile motion occurs when an object is launched into the air and moves under the influence of gravity.
2. The motion is two-dimensional, consisting of horizontal and vertical components.
3. Air resistance is often neglected in basic projectile motion problems.
4. The horizontal velocity $(v_x)$ remains constant since there is no acceleration in that direction.
5. The vertical velocity $(v_y)$ changes due to gravity $g = 9.8{m/s}^2$
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Range of a Projectile

6. The range is the horizontal distance a projectile travels before hitting the ground.
7. The formula for range $(R)$ is:
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   $R = \frac{u^2\sin(2\theta)}{g}$
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   where $u$ is the initial velocity, $\theta$ is the angle of projection, and $g$ is acceleration due to gravity.
8. The range is maximum when the angle of projection is $45^\circ$.
9. Range increases with greater initial velocity $(u)$.
10. The range is zero if the object is projected straight up $(\theta = 90^\circ)$.
11. For the same initial speed, complementary angles $(e.g., 30^\circ and 60^\circ)$ yield the same range.
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Maximum Height

12. The maximum height is the highest vertical position reached by the projectile.
13. The formula for maximum height $(H)$ is:
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    $H = \frac{u^2 \sin^2(\theta)}{2g}$
14. At maximum height, the vertical velocity $(v_y)$ becomes zero momentarily.
15. The greater the angle of projection, the higher the maximum height up to $90^\circ$.
16. Maximum height is independent of the horizontal velocity.
17. Increasing the initial velocity increases the maximum height.
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Time of Flight

18. The time of flight $(T)$ is the total time the projectile spends in the air.
19. The formula for time of flight is:
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    $T = \frac{2u \sin(\theta)}{g}$
20. Time of flight depends on the vertical component of the initial velocity $(u \sin(\theta))$.
21. Doubling the initial velocity doubles the time of flight.
22. Time of flight is longer for steeper angles of projection.
23. For vertical motion $(\theta = 90^\circ)$, the time of flight is maximum for the given velocity.
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Key Observations

24. Range, maximum height, and time of flight are interconnected.
25. If the launch angle is increased beyond $45^\circ$, the range decreases, but the height and time increase.
26. Horizontal and vertical motions are independent of each other.
27. At $t = T/2$, the projectile reaches maximum height.
28. The horizontal velocity $(v_x)$ at any time is $v_x = u \cos(\theta)$.
29. The vertical velocity $(v_y)$ at any time is:
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    $v_y = u \sin(\theta) - g t$
30. Displacement in the vertical direction $(y)$ is given by:
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    $y = u \sin(\theta) t - \frac{1}{2} g t^2$
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Applications of Projectile Motion

31. Sports: Understanding projectile motion helps in predicting the trajectory of balls in games like football, basketball, or cricket.
32. Space Exploration: Rockets follow projectile motion paths in their initial phases.
33. Artillery and Ballistics: Engineers use projectile motion principles to design and aim artillery.
34. Aviation: Pilots and engineers analyze projectile motion during takeoffs and landings.
35. Physics Experiments: Projectile motion is studied in labs to understand kinematics and dynamics.
36. Entertainment: Stunt coordinators in movies use projectile motion to plan action scenes.
37. Engineering: Architects use projectile motion concepts to design curved structures like arches.
38. Water Fountains: The motion of water in fountains follows a projectile trajectory.
39. Diving and Gymnastics: Athletes analyze projectile motion for precise landing and jumps.
40. Agriculture: Farmers use projectile motion principles in designing irrigation systems.
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Practical Considerations

41. Air resistance can reduce the range and maximum height in real-world situations.
42. In games like golf, wind direction affects the trajectory of the ball.
43. Launching objects on other planets requires accounting for different gravity values.
44. Parabolic paths in projectile motion are ideal; real-world trajectories deviate slightly.
45. In cannonball firing, elevation angles are adjusted to maximize range.
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Solving Projectile Motion Problems

46. Break the motion into horizontal and vertical components for simplicity.
47. Use $u \cos(\theta)$ for horizontal velocity and $u \sin(\theta)$ for vertical velocity.
48. Time of flight is calculated using vertical motion equations.
49. The range is computed using horizontal motion and time of flight.
50. Maximum height is determined using vertical motion equations.
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Jamb(utme) key points solving problems involving projectile motion

Question 1: Finding Time of Flight

1. A ball is projected vertically upwards with a velocity of $20m/s$. How long will it stay in the air?

• Solution:
• Total time of flight $(T)$: $T = \frac{2u}{g}$
• $T = \frac{2\times 20}{9.8} = 4.08s$
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Question 2: Maximum Height

2. What is the maximum height reached by the ball in Question 1?

• Solution:
• $H = \frac{u^2}{2g}$
• $H = \frac{20^2}{2 \times 9.8} = \frac{400}{19.6} = 20.41m$
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Question 3: Horizontal Range

3. A projectile is launched at $45^\circ$ with an initial velocity of $30m/s$. Find its range.

• Solution:
• $R = \frac{u^2 \sin(2\theta)}{g}$
• $R = \frac{30^2 \sin(90^\circ)}{9.8} = \frac{900}{9.8} = 91.84m$
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Question 4: Velocity at a Given Time

4. A projectile is launched at $60^\circ$ with $40m/s$. What is the vertical velocity after $2s$?

• Solution:
• $v_y = u \sin(\theta) - g t$
• $v_y = 40 \times 0.866 - 9.8 \times 2$
• $v_y = 34.64 - 19.6 = 15.04m/s$
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Question 5: Height at a Given Time

5. Find the height of the projectile in Question 4 after $2s$.

• Solution:
• $y = u \sin(\theta) t - \frac{1}{2} g t^2$
• $y = 40 \times 0.866 \times 2 - \frac{1}{2} \times 9.8 \times 4$
• $y = 69.28 - 19.6 = 49.68m$
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Question 6: Time to Reach Maximum Height

6. How long does it take to reach the maximum height for a projectile launched at $50m/s$ at $30^\circ$?

• Solution:
• $t_{\text{up}} = \frac{u \sin(\theta)}{g}$
• $t_{\text{up}} = \frac{50 \times 0.5}{9.8} = 2.55s$
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Question 7: Maximum Height

7. Find the maximum height for the projectile in Question 6.

• Solution:
• $H = \frac{u^2 \sin^2(\theta)}{2g}$
• $H = \frac{50^2 \times 0.25}{2 \times 9.8} = \frac{625}{19.6} = 31.89m$
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Question 8: Horizontal Displacement After Given Time

8. A projectile is launched at $40^\circ$ with $25m/s$. Find its horizontal displacement after $3s$.

• Solution:
• $x = u \cos(\theta) t$
• $x = 25 \times 0.766 \times 3 = 57.45m$
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Question 9: Total Time of Flight

9. Calculate the total time of flight for the projectile in Question 8.

• Solution:
• $T = \frac{2u \sin(\theta)}{g}$
• $T = \frac{2 \times 25 \times 0.643}{9.8} = 3.28s$
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Question 10: Range

10. Find the range of the projectile in Question 8.

• Solution:
• $R = u \cos(\theta) T$
• $R = 25 \times 0.766 \times 3.28 = 62.83m$
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Question 11: Finding Initial Velocity

11. A projectile is launched at $45^\circ$ and lands $100m$ away. Find its initial velocity.

• Solution:
• $u = \sqrt{\frac{R g}{\sin(2\theta)}}$
• $u = \sqrt{\frac{100 \times 9.8}{1}} = \sqrt{980} = 31.3m/s$
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Question 12: Angle for Maximum Range

12. At what angle should a projectile be launched to achieve maximum range?

• Solution:
• $\theta = 45^\circ$.
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Question 13: Projectile Fired from an Elevated Platform

13. A projectile is launched from a $10m$ height at $30^\circ$ with $20m/s$. How long is it in the air?

• Solution:
• Use $y = h + u \sin(\theta) t - \frac{1}{2} g t^2$.
• Solve the quadratic: $0 = 10 + 20 \times 0.5 t - 4.9 t^2$.
• $t \approx 3.1s$.
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Question 14: Range from Elevated Platform

14. Find the range of the projectile in Question 13.

• Solution:
• $R = u \cos(\theta) t$
• $R = 20 \times 0.866 \times 3.1 = 53.74m$
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Question 15: Impact Velocity

15. What is the velocity of the projectile in Question 13 upon impact?

• Solution:
• $v_x = u \cos(\theta) = 20 \times 0.866 = 17.32m/s$
• $v_y = u \sin(\theta) - g t = 20 \times 0.5 - 9.8 \times 3.1 = -10.38m/s$
• $v = \sqrt{v_x^2 + v_y^2} = \sqrt{17.32^2 + (-10.38)^2} = 20m/s$.
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Question 16: Maximum Height from Elevated Platform

16. A projectile is launched from a height of $15m$ at $60^\circ$ with $25m/s$. Find its maximum height.

• Solution:
• Height above the platform:
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  $H = \frac{u^2 \sin^2(\theta)}{2g}$ $H = \frac{25^2 \times (0.866)^2}{2 \times 9.8} = \frac{469.8}{19.6} = 23.96m$.
  • Total height: $H_{\text{total}} = H + 15 = 23.96 + 15 = 38.96 \, \text{m}.$
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Question 17: Horizontal Displacement After 2 Seconds

17. A projectile is launched at $50^\circ$ with $30m/s$. Find the horizontal displacement after $2s$.

• Solution:
  • $x = u \cos(\theta) t$
  • $x = 30 \times 0.6428 \times 2 = 38.57m$.
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Question 18: Final Velocity of Projectile

18. A projectile is launched at $45^\circ$ with $40m/s$. Find the velocity upon hitting the ground.

• Solution:
  • $v_x = u \cos(\theta) = 40 \times 0.707 = 28.28m/s$, $v_y = u \sin(\theta) = 40 \times 0.707 = 28.28m/s$.
  • Final velocity magnitude: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{28.28^2 + 28.28^2} = 40m/s.$
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Question 19: Range of a Projectile on a Cliff

19. A projectile is launched horizontally from a $50m$ high cliff at $15m/s$. Find the range.

• Solution:
  • Time to hit the ground: $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 50}{9.8}} = 3.19s.$
  • Range: $R = u_x t = 15 \times 3.19 = 47.85m.$
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Question 20: Time of Flight for Object Thrown Downward

20. An object is thrown downward from a $20m$ height with $10m/s$. Find the time to hit the ground.

• Solution:
  • Solve $y = h + u t + \frac{1}{2} g t^2$: $20 = 10 t + 4.9 t^2.$
    • Using the quadratic formula, $t = 1.27s$.
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Question 21: Time to Fall Back to Launch Height

21. A projectile is launched at $70^\circ$ with $40m/s$. How long does it take to return to the launch height?

• Solution:
  • Total time: $T = \frac{2u \sin(\theta)}{g}.$ $T = \frac{2 \times 40 \times 0.94}{9.8} = 7.67s$
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Question 22: Horizontal Velocity After Given Time

22. What is the horizontal velocity of the projectile in Question 21 after $4s$?

• Solution:
  • Horizontal velocity remains constant: $v_x = u \cos(\theta) = 40 \times 0.34 = 13.6m/s.$
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Question 23: Total Horizontal Range

23. Find the total horizontal range for the projectile in Question 21.

• Solution:
  • $R = u \cos(\theta) T$: $R = 40 \times 0.34 \times 7.67 = 104.24m.$
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Question 24: Maximum Height for Horizontal Launch

24. A ball is thrown horizontally at $20m/s$ from a $40m$ high building. What is its maximum height above the ground?

• Solution:
  • Maximum height is the initial height: $H = 4m.$
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Question 25: Impact Velocity for Horizontal Launch

25. Find the velocity of the ball in Question 24 upon impact.

• Solution:
  • $v_x = 20m/s$, $v_y = \sqrt{2 g h} = \sqrt{2 \times 9.8 \times 40} = 28m/s$.
  • Final velocity: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 28^2} = 34.41m/s.$
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Question 26: Angle of Impact

26. Find the angle of impact for the ball in Question 24.

• Solution:
  • $\tan(\theta) = \frac{v_y}{v_x} = \frac{28}{20} = 1.4$.
  • $\theta = \tan^{-1}(1.4) = 54.46^\circ$.
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Question 27: Time of Flight for Level Ground

27. A projectile is launched at $25^\circ$ with $50m/s$. Find the time of flight.

• Solution:
  • $T = \frac{2u \sin(\theta)}{g}$, $T = \frac{2 \times 50 \times 0.4226}{9.8} = 4.31 \, \text{s}.$
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Question 28: Range for Level Ground

28. Find the range for the projectile in Question 27.

• Solution:
  • $R = u \cos(\theta) T$: $R = 50 \times 0.9063 \times 4.31 = 195.3m.$
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Question 29: Maximum Height for Level Ground

29. Find the maximum height for the projectile in Question 27.

• Solution:
  • $H = \frac{u^2 \sin^2(\theta)}{2g}$: $H = \frac{50^2 \times 0.1785}{19.6} = 22.8 \, \text{m}.$
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Question 30: Range with Increased Initial Speed

30. If the speed in Question 27 increases to $60m/s$, find the new range.

• Solution:
  • $R = \frac{u^2 \sin(2\theta)}{g}$: $R = \frac{60^2 \times 0.766}{9.8} = 281.7m.$

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