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.
1 answer