To find the horizontal distance (denoted as $x$) that the stone travels before hitting the ground, we need to break down the problem into horizontal and vertical components. This is a projectile motion problem.

Step 1: Decompose the initial velocity into components

The initial velocity ($v_0 = 43.3 \, \text{m/s}$) is launched at an angle of $30^\circ$.

• Horizontal velocity ($v_{0x}$): $v_{0x} = v_0 \cos(\theta) = 43.3 \cos(30^\circ)$ $v_{0x} = 43.3 \times \frac{\sqrt{3}}{2} = 37.5 \, \text{m/s}$

• Vertical velocity ($v_{0y}$): $v_{0y} = v_0 \sin(\theta) = 43.3 \sin(30^\circ)$ $v_{0y} = 43.3 \times \frac{1}{2} = 21.65 \, \text{m/s}$

Step 2: Time of flight

To find the time it takes for the stone to hit the ground, we use the vertical motion. The vertical displacement is $y = -164 \, \text{m}$ (since it falls 164 m from the building top).

Using the kinematic equation for vertical motion: $y = v_{0y} t + \frac{1}{2} a t^2$ where:

• $y = -164 \, \text{m}$
• $v_{0y} = 21.65 \, \text{m/s}$
• $a = -9.8 \, \text{m/s}^2$ (acceleration due to gravity)
• $t = ?$ (time of flight)

Substitute the known values: $-164 = 21.65 t - \frac{1}{2} (9.8) t^2$ $-164 = 21.65 t - 4.9 t^2$ Rearranging: $4.9 t^2 - 21.65 t - 164 = 0$

Now, solve this quadratic equation for $t$ using the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 4.9$, $b = -21.65$, and $c = -164$.

Step 3: Find horizontal distance $x$

Once we calculate $t$, we can find the horizontal distance using the formula: $x = v_{0x} t$ where $v_{0x} = 37.5 \, \text{m/s}$ and $t$ is the time of flight we just found.

Let me calculate the time $t$ and the distance $x$.The time of flight is approximately $8.40 \, \text{seconds}$, and the horizontal distance traveled (denoted as $x$) is approximately $315.06 \, \text{meters}$.

Would you like further details or have any questions?

Here are some related questions to consider:

1. How would changing the launch angle affect the horizontal distance?
2. What happens if the initial velocity is increased while keeping the same angle?
3. How would air resistance alter the result compared to the ideal projectile motion calculated here?
4. Can you calculate the maximum height reached by the stone?
5. What would the horizontal distance be if the stone was thrown from a lower height?

Tip: In projectile motion, the horizontal velocity remains constant because there is no horizontal acceleration (ignoring air resistance).