Let's break down the problem step-by-step using the principles of projectile motion.

1. First Launch (Cliff Height $H$)

• The cannonball is launched horizontally, so the initial vertical velocity ($v_{y0}$) is $0$.

• Vertical motion is influenced by gravity. The time ($t$) it takes for the cannonball to fall from the height $H$ is given by the equation: $H = \frac{1}{2} g t^2$ Solving for $t$: $t = \sqrt{\frac{2H}{g}}$

• Horizontal motion is constant since there's no horizontal acceleration. If $v_x$ is the horizontal velocity, the horizontal distance ($D$) covered is: $D = v_x t$ Substituting $t$: $D = v_x \sqrt{\frac{2H}{g}}$

2. Second Launch (Cliff Height $2H$)

• Using the same logic, for the second launch, the time ($t'$) it takes to fall from the height $2H$ is: $2H = \frac{1}{2} g t'^2$ Solving for $t'$: $t' = \sqrt{\frac{4H}{g}} = 2\sqrt{\frac{H}{g}}$

• The horizontal distance ($D'$) covered in this case, with the same horizontal velocity $v_x$, is: $D' = v_x t'$ Substituting $t'$: $D' = v_x \times 2\sqrt{\frac{H}{g}}$

3. Relationship Between Distances

From the first launch, we have $D = v_x \sqrt{\frac{2H}{g}}$. We can express $v_x$ in terms of $D$ and $H$: $v_x = \frac{D}{\sqrt{\frac{2H}{g}}}$

Now substitute $v_x$ into $D'$: $D' = \frac{D}{\sqrt{\frac{2H}{g}}} \times 2\sqrt{\frac{H}{g}}$

Simplifying this: $D' = D \times \sqrt{2}$

Conclusion

When the cannonball is launched from a cliff of height $2H$, it will land a horizontal distance of $D \times \sqrt{2}$ from the base of the cliff. If the original distance $D$ is given, then the new distance is approximately $1.41D$.

Would you like further clarification or more details on this?

Here are some related questions you might find interesting:

1. How does changing the launch angle affect the horizontal distance?
2. What if the cannonball was launched at an angle instead of horizontally?
3. How would air resistance impact the motion of the cannonball?
4. What is the maximum height achieved by a projectile launched at an angle?
5. How can you derive the time of flight for a projectile launched at an arbitrary angle?

Tip: In projectile motion, horizontal and vertical motions are independent of each other. Horizontal motion has constant velocity, while vertical motion is influenced by gravity.