Question

From the top of a tall cliff of height y, a soccer ball is kicked horizontally so that it leaves the cliff with a velocity v. Assuming air friction is negligible, the speed of the ball just before it hits the ground is:

Answer 1

Answer:

$\sqrt{v^{2} + 2\, g\, y}$.

Explanation:

Apply the SUVAT equation ${v_{1}}^{2} - {v_{0}}^{2} = 2\, a\, x$ to the vertical component of the soccer ball's motion to find the vertical velocity right before the ball lands. In this equation:

• $v_{1}$ represents the vertical velocity of the ball right before it lands (needs to be found);
• $v_{0}$ represents the initial vertical velocity of the soccer ball ($0$ in this question since the ball was initially travelling horizontally);
• $a$ represents the acceleration (equal to $g$ since air friction is negligible);
• $x$ represents the vertical displacement of this ball (in this question, this vertical displacement is equal $y$, the height of the cliff.)

Rearrange this equation to find an expression for the vertical velocity $v_{1}$ in terms of $y$ and $g$:

$\begin{aligned}v_{1} &= \sqrt{{v_{0}}^{2} + 2\, a\, x} \\ &= \sqrt{2\, g\, y}\end{aligned}$.

Since air friction is negligible, the horizontal velocity of the ball would stay unchanged. Thus, the horizontal velocity of the ball would still be $v$ right before the ball lands.

Notice that these two components of velocity are perpendicular to one another. Thus, find the combined velocity of the ball right before landing could be found using Pythagorean's Theorem:

$\sqrt{v^{2} + {v_{1}}^{2}} = \sqrt{v^{2} + 2\, g\, y}$.