A heavy object is thrown horizontally from a 23 m high cliff. It strikes the ground 73 m from the base of the cliff. At what speed was it thrown?

To determine the speed at which the heavy object was thrown horizontally from the cliff, we can analyze the motion in two dimensions: vertical and horizontal.

### Step 1: Analyze the vertical motion

The vertical motion is influenced by gravity. The formula for the vertical distance \( h \) fallen after time \( t \) is given by:

\[
h = \frac{1}{2} g t^2
\]

Where:
- \( h = 23 \) m (the height of the cliff)
- \( g \approx 9.81 \) m/s² (acceleration due to gravity)

Rearranging this formula to find time \( t \) gives us:

\[
t^2 = \frac{2h}{g}
\]
\[
t = \sqrt{\frac{2h}{g}}
\]

Plugging in the values:

\[
t = \sqrt{\frac{2 \times 23}{9.81}} \approx \sqrt{\frac{46}{9.81}} \approx \sqrt{4.688} \approx 2.17 \text{ s}
\]

### Step 2: Analyze the horizontal motion

The horizontal distance \( d \) traveled is given by the formula:

\[
d = v_x t
\]

Where:
- \( d = 73 \) m (the horizontal distance from the base of the cliff)
- \( v_x \) is the horizontal speed, which we need to find.

Rearranging the formula to solve for \( v_x \):

\[
v_x = \frac{d}{t}
\]

Substituting the values into the equation gives:

\[
v_x = \frac{73}{2.17} \approx 33.65 \text{ m/s}
\]

### Conclusion

The speed at which the heavy object was thrown horizontally is approximately **33.65 m/s**.