Q.A javelin is released with a speed of 14 m/s. Taking g = 9.8 m/s² and using the level-ground range formula R = (v² sin 2θ)/g (air resistance neglected),

This is a numerical question requiring calculations based on a given formula, followed by a conceptual interpretation.

(i)

To calculate the range for an angle of projection of 45^°, we use the given formula:

Given values:

v = 14 m/s

θ = 45^°

g = 9.8 m/s²

Substitute these values into the formula:

R = (14 m/s)² sin (2 × 45^°)/(9.8 m/s²)

R = (196 m²/s² × sin (90^°))/(9.8 m/s²)

Since sin 90^° = 1:

R = (196 m²/s² × 1)/(9.8 m/s²)

R = 196/9.8 m

R = 20 m

(ii)

To calculate the range for an angle of projection of 30^°, we use the same formula:

Given values:

v = 14 m/s

θ = 30^°

g = 9.8 m/s²

Substitute these values into the formula:

R = (14 m/s)² sin (2 × 30^°)/(9.8 m/s²)

R = (196 m²/s² × sin (60^°))/(9.8 m/s²)

Since sin 60^° = √(3)/2 ≈ 0.866:

R = (196 m²/s² × 0.866)/(9.8 m/s²)

R = 169.736/9.8 m

R ≈ 17.32 m

(iii) …

Understanding projectile motion is fundamental in many sports, especially throwing events like javelin, shot put, and discus. The path an object takes after being thrown is called its trajectory, and the horizontal distance it covers before landing is its range. Factors like the initial speed, the angle at which it is thrown, and gravity all play a crucial role in determining this range. For an athlete, optimizing these factors is key to achieving maximum distance.

The formula provided for the range of a projectile launched and landing at the same height, neglecting air resistance, is:

Where:

• R is the horizontal range (distance covered).
• v is the initial speed of projection.
• θ is the angle of projection with respect to the horizontal.
• g is the acceleration due to gravity.

We are given:

• Initial speed, v = 14 m/s
• Acceleration due to gravity, g = 9.8 m/s²

(i) Calculate the range for an angle of projection of 45°

1. Substitute the given values into the range formula for θ = 45^°:

   R = (14 m/s)² sin (2 × 45^°)/(9.8 m/s²)

2. Calculate v² and 2θ:

   R = (196 m²/s² sin (90^°))/(9.8 m/s²)

3. The value of sin(90^°) is 1:

   R = (196 m²/s² × 1)/(9.8 m/s²)

4. Perform the division:

   R = 196/9.8 m

   R = 20 m

(ii) Calculate the range for an angle of projection of 30°

1. Substitute the given values into the range formula for θ = 30^°:

   R = (14 m/s)² sin (2 × 30^°)/(9.8 m/s²)

2. Calculate v² and 2θ:

   R = (196 m²/s² sin (60^°))/(9.8 m/s²)

3. The value of sin(60^°) is √(3)/2 ≈ 0.8660:

   R = (196 m²/s² × 0.8660)/(9.8 m/s²)

4. Perform the division:

   R = 20 × 0.8660 m

   R ≈ 17.32 m

(iii) Comment on what your two answers show about the effect of the angle of projection on the distance covered …