17) An object is thrown directly downward from the top of a 180-meter-tall building. It takes 1.0 seconds for the object to fall the last 60 meters. With what initial downward speed was the object thrown from the roof?

A) 15 m/s
B) 25 m/s
C) 35 m/s
D) 55 m/s
E) insufficient information
1. Let's denote the initial speed of the object as \( v_i \), the acceleration due to gravity as \( a = 10 \, \text{m/s}^2 \), and the height of the building as \( h = 180 \, \text{m} \).

2. We know that the object takes 1.0 seconds to fall the last 60 meters. We can use the kinematic equation for the distance traveled under constant acceleration:
   \[
   \Delta x = v_i t + \frac{1}{2} a t^2
   \]
   Here, \(\Delta x = 60 \, \text{m}\), \(t = 1.0 \, \text{s}\), and \(a = 10 \, \text{m/s}^2\).

3. Plugging in the values, we get:
   \[
   60 = v_i (1) + \frac{1}{2} (10) (1)^2
   \]
   Simplifying this, we get:
   \[
   60 = v_i + 5
   \]
   Solving for \( v_i \), we find:
   \[
   v_i = 55 \, \text{m/s}
   \]
   This is the velocity of the object 60 meters above the ground.

4. Next, we need to find the initial speed of the object when it was thrown from the roof. We use the kinematic equation relating the final velocity, initial velocity, acceleration, and distance:
   \[
   v_f^2 = v_i^2 + 2a\Delta x
   \]
   Here, \( v_f = 55 \, \text{m/s} \), \( a = 10 \, \text{m/s}^2 \), and \( \Delta x = 120 \, \text{m} \) (since the object falls 120 meters from the roof to the point where it is 60 meters above the ground).

5. Plugging in the values, we get:
   \[
   55^2 = v_i^2 + 2 \cdot 10 \cdot 120
   \]
   Simplifying this, we get:
   \[
   3025 = v_i^2 + 2400
   \]
   Solving for \( v_i^2 \), we find:
   \[
   v_i^2 = 3025 - 2400
   \]
   \[
   v_i^2 = 625
   \]
   Taking the square root of both sides, we get:
   \[
   v_i = 25 \, \text{m/s}
   \]

The final answer is \(\boxed{25 \, \text{m/s}}\).