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: $$25=\frac{1}{2}(9.81)t^2$$ $$25=4.905t^2$$ 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 $19.91,\text{m/s}$.