Projectile Motion Answer Key

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Date:

📚 Part 1: Multiple Choice Questions

1. What is the acceleration of a projectile in the horizontal direction (ignoring air resistance)?

✓

0 m/s²

9.8 m/s² downward

9.8 m/s² upward

Variable depending on initial velocity

Answer: 0 m/s² - There are no forces acting horizontally (ignoring air resistance), so horizontal acceleration is zero.

2. At what point in a projectile's trajectory is the vertical velocity zero?

At launch

✓

At maximum height

At landing

Never

Answer: At maximum height - The projectile momentarily stops moving vertically before falling back down.

3. Which factors affect the range of a projectile? (Select all that apply)

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Initial velocity

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Launch angle

Mass of the projectile

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Height of launch

Answer: Initial velocity, launch angle, and height of launch all affect range. Mass does not affect range (ignoring air resistance).

4. What is the optimal launch angle for maximum range on level ground?

30°

✓

45°

60°

90°

Answer: 45° - This angle provides the optimal balance between horizontal and vertical velocity components for maximum range.

🔢 Part 2: Calculations and Problem Solving

5. A ball is thrown horizontally from a 20 m high building with an initial velocity of 15 m/s. Calculate the time of flight.

Given: h = 20 m, v₀ₓ = 15 m/s, v₀ᵧ = 0 m/s, g = 9.8 m/s²

Formula: h = v₀ᵧt + ½gt²

Calculation: 20 = 0 + ½(9.8)t²

20 = 4.9t²

t² = 20/4.9 = 4.08

Answer: t = 2.02 seconds

6. Using the data from Question 5, calculate the horizontal distance travelled.

Formula: x = v₀ₓ × t

Calculation: x = 15 × 2.02

Answer: x = 30.3 metres

7. A projectile is launched at 25 m/s at an angle of 30° above horizontal. Calculate the horizontal and vertical components of initial velocity.

Given: v₀ = 25 m/s, θ = 30°

Horizontal component: v₀ₓ = v₀ cos θ = 25 × cos(30°) = 25 × 0.866 = 21.7 m/s

Vertical component: v₀ᵧ = v₀ sin θ = 25 × sin(30°) = 25 × 0.5 = 12.5 m/s

✏️ Part 3: Short Answer Questions

8. Explain why the horizontal velocity of a projectile remains constant (ignoring air resistance).

Answer: The horizontal velocity remains constant because there are no horizontal forces acting on the projectile (ignoring air resistance). According to Newton's First Law, an object in motion stays in motion at constant velocity unless acted upon by an external force. Only gravity acts on the projectile, and it acts vertically downward, not horizontally.

9. Describe what happens to the trajectory of a projectile if the launch angle is increased from 30° to 60°, assuming the same initial speed.

Answer: Increasing the launch angle from 30° to 60° will result in a higher maximum height but a shorter horizontal range. The projectile will spend more time in the air but travel less distance horizontally because more of the initial velocity is directed upward rather than forward. The trajectory becomes more curved and peaked.

10. Draw and label a diagram showing the velocity vectors at three points during projectile motion: launch, maximum height, and just before landing.

Answer: Students should draw a parabolic trajectory with:

• At launch: Velocity vector pointing diagonally upward (both horizontal and vertical components)

• At maximum height: Velocity vector pointing horizontally (only horizontal component, vertical = 0)

• Before landing: Velocity vector pointing diagonally downward (horizontal component same as launch, vertical component equal and opposite to launch)