A ball is projected at an angle 30° above with the horizontal from the top of a tower and strikes the ground in 5 sec at an angle 45° with horizontal .find height of tower and the speed with which it was projected

To solve this problem, we need to break it down into manageable parts using the principles of projectile motion. We know that the ball is projected from a height and strikes the ground after 5 seconds at a specific angle. Let's go through the calculations step by step.

Understanding the Motion

The ball is projected at an angle of 30° above the horizontal. The time of flight is 5 seconds, and it hits the ground at an angle of 45°. We can use these details to find both the height of the tower and the initial speed of the ball.

Step 1: Analyzing the Final Velocity

When the ball strikes the ground at a 45° angle, we can infer that the horizontal and vertical components of its velocity are equal. Let's denote the final velocity components as \( V_{fx} \) and \( V_{fy} \). Since the angle is 45°, we have:

• \( V_{fx} = V_{fy} \)

Step 2: Calculating the Final Velocity Components

The vertical component of the velocity can be calculated using the formula:

• \( V_{fy} = V_{iy} - g \cdot t \)

Where \( V_{iy} \) is the initial vertical velocity, \( g \) is the acceleration due to gravity (approximately 9.81 m/s²), and \( t \) is the time of flight (5 seconds). The initial vertical velocity can be expressed as:

• \( V_{iy} = V_i \cdot \sin(30°) \)

Substituting this into the equation gives:

• \( V_{fy} = V_i \cdot \sin(30°) - g \cdot 5 \)

Step 3: Finding the Initial Velocity

Since \( V_{fx} \) is equal to \( V_{fy} \), we can express the horizontal component as:

• \( V_{fx} = V_i \cdot \cos(30°) \)

Setting \( V_{fx} \) equal to \( V_{fy} \) gives us:

• \( V_i \cdot \cos(30°) = V_i \cdot \sin(30°) - g \cdot 5 \)

Now, substituting the values of \( \sin(30°) = 0.5 \) and \( \cos(30°) = \sqrt{3}/2 \), we can simplify the equation:

• \( V_i \cdot \frac{\sqrt{3}}{2} = V_i \cdot 0.5 - 9.81 \cdot 5 \)

Rearranging gives:

• \( V_i \cdot \frac{\sqrt{3}}{2} - V_i \cdot 0.5 = -49.05 \)

Factoring out \( V_i \) leads to:

• \( V_i \left( \frac{\sqrt{3}}{2} - 0.5 \right) = -49.05 \)

Calculating \( \frac{\sqrt{3}}{2} - 0.5 \) gives approximately 0.366. Thus:

• \( V_i \cdot 0.366 = 49.05 \)

Solving for \( V_i \) yields:

• \( V_i \approx \frac{49.05}{0.366} \approx 134.5 \, \text{m/s} \)

Step 4: Finding the Height of the Tower

Now that we have the initial velocity, we can find the height of the tower using the vertical motion equation:

• \( h = V_{iy} \cdot t - \frac{1}{2} g t^2 \)

Substituting \( V_{iy} = V_i \cdot \sin(30°) = 134.5 \cdot 0.5 = 67.25 \, \text{m/s} \):

• \( h = 67.25 \cdot 5 - \frac{1}{2} \cdot 9.81 \cdot 5^2 \)

Calculating gives:

• \( h = 336.25 - 122.625 = 213.625 \, \text{m} \)

Final Results

In summary, the height of the tower is approximately 213.63 meters, and the speed with which the ball was projected is about 134.5 m/s. This example illustrates how projectile motion can be analyzed using both horizontal and vertical components to derive meaningful results.