Question

give me some solved examples of projectile motion

Answer 1

Here are some solved examples of projectile motion:

Answer 2

Here are some solved examples of projectile motion:

Solved Example 1: Basic Projectile Motion Parameters

Question: A projectile is fired with an initial velocity of 20 m/s at an angle of 30° above the horizontal. Calculate its time of flight, maximum height, and horizontal range. (Assume g = 10 m/s²)

Solution:

Given:

• Initial velocity, $u = 20 \, \text{m/s}$
• Angle of projection, $\theta = 30^\circ$
• Acceleration due to gravity, $g = 10 \, \text{m/s}^2$

Formulas Used:

• Time of Flight ($T$): $T = \frac{2u \sin\theta}{g}$
• Maximum Height ($H$): $H = \frac{u^2 \sin^2\theta}{2g}$
• Horizontal Range ($R$): $R = \frac{u^2 \sin(2\theta)}{g}$

Calculations:

1. Time of Flight (T): $T = \frac{2 \times 20 \times \sin(30^\circ)}{10} = \frac{40 \times 0.5}{10} = \frac{20}{10} = 2 \, \text{s}$

2. Maximum Height (H): $H = \frac{(20)^2 \times (\sin(30^\circ))^2}{2 \times 10} = \frac{400 \times (0.5)^2}{20} = \frac{400 \times 0.25}{20} = \frac{100}{20} = 5 \, \text{m}$

3. Horizontal Range (R): $R = \frac{(20)^2 \times \sin(2 \times 30^\circ)}{10} = \frac{400 \times \sin(60^\circ)}{10} = \frac{400 \times \frac{\sqrt{3}}{2}}{10} = \frac{200\sqrt{3}}{10} = 20\sqrt{3} \approx 34.64 \, \text{m}$

Explanation: The standard formulas for time of flight, maximum height, and horizontal range were directly applied using the given initial velocity, projection angle, and acceleration due to gravity.

Answer: Time of Flight = 2 s Maximum Height = 5 m Horizontal Range = 34.64 m

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Solved Example 2: Finding Initial Velocity and Angle

Question: A cricket ball is thrown such that it reaches a maximum height of 5 m and has a horizontal range of 20 m. Find the initial velocity and angle of projection. (Assume g = 10 m/s²)

Solution:

Given:

• Maximum Height, $H = 5 \, \text{m}$
• Horizontal Range, $R = 20 \, \text{m}$
• Acceleration due to gravity, $g = 10 \, \text{m/s}^2$

Formulas Used:

• $H = \frac{u^2 \sin^2\theta}{2g}$
• $R = \frac{u^2 \sin(2\theta)}{g}$
• Relationship: $\tan\theta = \frac{4H}{R}$

Calculations:

1. Angle of Projection ($\theta$): Using the relationship between H and R: $\tan\theta = \frac{4H}{R} = \frac{4 \times 5}{20} = \frac{20}{20} = 1$ Therefore, $\theta = \tan^{-1}(1) = 45^\circ$.

2. Initial Velocity (u): Using the horizontal range formula: $R = \frac{u^2 \sin(2\theta)}{g}$ $20 = \frac{u^2 \sin(2 \times 45^\circ)}{10}$ $20 = \frac{u^2 \sin(90^\circ)}{10}$ $20 = \frac{u^2 \times 1}{10}$ $u^2 = 200$ $u = \sqrt{200} = 10\sqrt{2} \, \text{m/s} \approx 14.14 \, \text{m/s}$

Explanation: The angle of projection was found using the direct relationship between maximum height and horizontal range ($\tan\theta = \frac{4H}{R}$). Once the angle was determined, the initial velocity was calculated using the horizontal range formula.

Answer: Initial velocity = $10\sqrt{2}$ m/s (or approx. 14.14 m/s) Angle of projection = 45°

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Solved Example 3: Projectile Fired Horizontally from a Height

Question: A stone is thrown horizontally from the top of a 45 m high building with a speed of 10 m/s. How far from the base of the building will the stone strike the ground? (Assume g = 10 m/s²)

Solution:

Given:

• Initial horizontal velocity, $u_x = 10 \, \text{m/s}$
• Initial vertical velocity, $u_y = 0 \, \text{m/s}$ (since thrown horizontally)
• Height of the building, $H = 45 \, \text{m}$
• Acceleration due to gravity, $g = 10 \, \text{m/s}^2$

Formulas Used:

• Vertical motion: $s_y = u_y t + \frac{1}{2} a_y t^2$
• Horizontal motion: $s_x = u_x t$

Calculations:

1. Time of Flight (T): Consider vertical motion. The stone falls a height of 45 m. $H = u_y T + \frac{1}{2} g T^2$ $45 = 0 \times T + \frac{1}{2} \times 10 \times T^2$ $45 = 5 T^2$ $T^2 = \frac{45}{5} = 9$ $T = \sqrt{9} = 3 \, \text{s}$ (Time cannot be negative)

2. Horizontal Distance (Range, R): Consider horizontal motion. The horizontal velocity remains constant. $R = u_x \times T$ $R = 10 \, \text{m/s} \times 3 \, \text{s}$ $R = 30 \, \text{m}$

Explanation: The problem was solved by first determining the time it takes for the stone to hit the ground using the equations of motion for vertical displacement. Since the horizontal velocity remains constant in projectile motion (neglecting air resistance), this time was then used to calculate the horizontal distance covered.

Answer: The stone will strike the ground 30 m from the base of the building.