a stone is thrown towards a pole of a height 15m and at a distance of 30m. the stone passes just over the pole going downwards, making angle 45 with horizontal. the speed v of stone is

To determine the speed of the stone when it is thrown towards the pole, we can use some principles from physics, particularly the equations of motion. The scenario describes a projectile motion where the stone is thrown at an angle and passes just over a pole of height 15 meters, located 30 meters away. Let's break this down step by step.

Understanding the Problem

We know the following:

• The height of the pole (h) = 15 m
• The horizontal distance to the pole (d) = 30 m
• The angle of descent when passing the pole (θ) = 45 degrees

Since the stone makes a 45-degree angle with the horizontal as it descends, we can use this information to find the initial speed (v) of the stone.

Breaking Down the Motion

In projectile motion, the horizontal and vertical components of the motion can be analyzed separately. The stone's speed can be broken down into two components:

• Horizontal component: \( v_x = v \cdot \cos(θ) \)
• Vertical component: \( v_y = v \cdot \sin(θ) \)

At a 45-degree angle, both components are equal, so we can say:

\( v_x = v_y = \frac{v}{\sqrt{2}} \)

Using the Time of Flight

To find the speed, we first need to determine the time it takes for the stone to reach the pole. The horizontal motion can be described by the equation:

\( d = v_x \cdot t \)

Substituting for \( v_x \):

\( 30 = \frac{v}{\sqrt{2}} \cdot t \)

From this, we can express time (t) as:

\( t = \frac{30 \sqrt{2}}{v} \)

Vertical Motion Analysis

For the vertical motion, we can use the equation of motion:

\( h = v_y \cdot t - \frac{1}{2} g t^2 \)

Substituting for \( v_y \) and \( t \):

\( 15 = \frac{v}{\sqrt{2}} \cdot \left(\frac{30 \sqrt{2}}{v}\right) - \frac{1}{2} g \left(\frac{30 \sqrt{2}}{v}\right)^2 \)

Now, simplifying this equation:

\( 15 = 30 - \frac{1}{2} g \cdot \frac{1800}{v^2} \)

Where \( g \) (acceleration due to gravity) is approximately 9.81 m/s². Rearranging gives:

\( \frac{1}{2} g \cdot \frac{1800}{v^2} = 30 - 15 \)

\( \frac{1}{2} g \cdot \frac{1800}{v^2} = 15 \)

Multiplying both sides by \( v^2 \) and rearranging gives:

\( 900g = 15v^2 \)

Thus, we can solve for \( v^2 \):

\( v^2 = \frac{900g}{15} \)

\( v^2 = 60g \)

Substituting \( g \approx 9.81 \):

\( v^2 = 60 \cdot 9.81 \approx 588.6 \)

Finally, taking the square root gives:

\( v \approx \sqrt{588.6} \approx 24.2 \, \text{m/s} \)

Final Result

The speed of the stone when it is thrown towards the pole is approximately 24.2 m/s. This calculation illustrates how projectile motion can be analyzed using both horizontal and vertical components, leading to a clear understanding of the stone's trajectory.