Question

A ball is thrown horizontally from a 25-meter cliff. The ball lands 45 meters away from the base of the cliff. Determine the initial velocity of the ball.

Answer 1

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 \( 19.91 \, \text{m/s} \).