To solve this problem, we can use the principles of projectile motion. Since the stone is projected at an angle and reaches the tower and the horizontal plane, we need to apply the kinematic equations that describe projectile motion in both the horizontal and vertical directions.

Step 1: Breakdown of the motion components

Let:

• $u$ be the initial velocity of projection.
• $\theta$ be the angle of projection.
• The horizontal distance between the point of projection and the tower is 144 m.
• The height of the tower is 12 m.
• The stone strikes the horizontal plane at a distance of 9.6 m beyond the tower, so the total horizontal distance from the point of projection is $144 + 9.6 = 153.6$ m.

Step 2: Horizontal motion

The horizontal motion of the stone is uniform (no acceleration), so we can use the equation for horizontal displacement:

$x = u \cos(\theta) \cdot t$

where:

• $x = 153.6$ m (total horizontal distance).
• $u \cos(\theta)$ is the horizontal velocity component.
• $t$ is the time of flight.

Step 3: Vertical motion

For vertical motion, the stone is subject to gravity, so we use the following kinematic equation for displacement in the vertical direction:

$y = u \sin(\theta) \cdot t - \frac{1}{2} g t^2$

where:

• $y = 12$ m (the height of the tower).
• $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity.
• $u \sin(\theta)$ is the vertical velocity component.

Step 4: Time to reach the tower

At the point where the stone clears the tower, the horizontal distance covered is 144 m, and the height of the stone is 12 m. Let's use the horizontal distance and vertical motion equations to solve for time $t_1$ when the stone reaches the tower.

From the horizontal motion equation:

$144 = u \cos(\theta) \cdot t_1$

So,

$t_1 = \frac{144}{u \cos(\theta)}$

Now substitute this into the vertical motion equation. At this time $t_1$, the height $y$ is 12 m:

$12 = u \sin(\theta) \cdot t_1 - \frac{1}{2} g t_1^2$

Substitute $t_1 = \frac{144}{u \cos(\theta)}$ into this equation:

$12 = u \sin(\theta) \cdot \frac{144}{u \cos(\theta)} - \frac{1}{2} g \left( \frac{144}{u \cos(\theta)} \right)^2$

Simplify:

$12 = 144 \tan(\theta) - \frac{1}{2} g \cdot \frac{144^2}{u^2 \cos^2(\theta)}$

Step 5: Time to hit the ground

The stone hits the ground at a total horizontal distance of 153.6 m. So, using the horizontal distance equation again, we have:

$153.6 = u \cos(\theta) \cdot t_2$

Thus,

$t_2 = \frac{153.6}{u \cos(\theta)}$

Substitute this into the vertical motion equation. At time $t_2$, the height $y$ is 0:

$0 = u \sin(\theta) \cdot t_2 - \frac{1}{2} g t_2^2$

Substitute $t_2 = \frac{153.6}{u \cos(\theta)}$:

$0 = u \sin(\theta) \cdot \frac{153.6}{u \cos(\theta)} - \frac{1}{2} g \cdot \frac{153.6^2}{u^2 \cos^2(\theta)}$

Simplify:

$0 = 153.6 \tan(\theta) - \frac{1}{2} g \cdot \frac{153.6^2}{u^2 \cos^2(\theta)}$

Now, you have two equations involving $\tan(\theta)$ and $u$. By solving these, we can find the values of $u$ and $\theta$.

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