To determine the initial velocity of the ball when it is thrown horizontally from the cliff, we can analyze the motion in two dimensions: vertical and horizontal.
Step 1: Analyze Vertical Motion
The vertical motion can be described using the following kinematic equation: \[ y = \frac{1}{2} g t^2 \] where:
- \( y \) is the vertical distance fallen (25 meters),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 , \text{m/s}^2 \)),
- \( t \) is the time in seconds.
Setting \( y = 25 \) meters, we have: \[ 25 = \frac{1}{2} (9.81) t^2 \] \[ 25 = 4.905 t^2 \] Now solve for \( t^2 \): \[ t^2 = \frac{25}{4.905} \approx 5.094 \] Taking the square root gives us: \[ t \approx \sqrt{5.094} \approx 2.26 , \text{seconds} \]
Step 2: Analyze Horizontal Motion
In horizontal motion, the distance \( x \) traveled by the ball is given by: \[ x = v_0 t \] where:
- \( x \) is the horizontal distance (45 meters),
- \( v_0 \) is the initial horizontal velocity (which we want to find),
- \( t \) is the time of flight (2.26 seconds).
Now substituting in the known values: \[ 45 = v_0 (2.26) \] Solving for \( v_0 \): \[ v_0 = \frac{45}{2.26} \approx 19.91 , \text{m/s} \]
Conclusion
The initial velocity of the ball when it is thrown horizontally from the cliff is approximately \( 19.91 , \text{m/s} \).
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