To solve this problem, we can break down the motion of the object into horizontal and vertical components.
Given:
- Initial speed of the object \( v_0 = 2000 \, \text{cm/s} \)
- Angle of projection \( \theta = 30^\circ \)
- Height of the building \( h = 45 \, \text{m} = 4500 \, \text{cm} \)
We'll start by analyzing the vertical motion to determine the time taken by the object to reach the ground.
1. **Vertical Motion:**
- The vertical component of the initial velocity is \( v_{0y} = v_0 \sin \theta \).
- We use the equation of motion: \( h = v_{0y} t - \frac{1}{2} g t^2 \), where \( g \) is the acceleration due to gravity.
Using \( v_{0y} = v_0 \sin \theta \), we get:
\[ 4500 \, \text{cm} = (2000 \, \text{cm/s}) \times (\sin 30^\circ) \times t - \frac{1}{2} \times (981 \, \text{cm/s}^2) \times t^2 \]
Solving for \( t \), we find the time taken for the object to reach the ground.
2. **Horizontal Motion:**
- The horizontal component of the initial velocity is \( v_{0x} = v_0 \cos \theta \).
- We use the equation \( d = v_{0x} \times t \) to find the horizontal distance traveled by the object.
We can now calculate the horizontal distance \( d \).
Let's solve for \( t \) first:
\[ 4500 = (2000 \times 0.5) \times t - \frac{1}{2} \times (981) \times t^2 \]
\[ 4500 = 1000t - 490.5t^2 \]
\[ 490.5t^2 - 1000t + 4500 = 0 \]
We can use the quadratic formula to solve for \( t \):
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where:
- \( a = 490.5 \)
- \( b = -1000 \)
- \( c = 4500 \)
Plugging in the values, we get:
\[ t = \frac{-(-1000) \pm \sqrt{(-1000)^2 - 4 \times 490.5 \times 4500}}{2 \times 490.5} \]
\[ t = \frac{1000 \pm \sqrt{1000000 - 8829000}}{981} \]
\[ t = \frac{1000 \pm \sqrt{-7829000}}{981} \]
Since the term under the square root is negative, it indicates that the object doesn't hit the ground.
There seems to be an inconsistency in the given problem. The object should strike the ground, but the calculation shows otherwise. Please review the problem statement and values provided.
