# How long should the ball take to fall on the floor from the time it leaves the edge of the table...?

## A golf ball is rolling off a table top w/c is 1.0m above the floor. The golf ball has an initial horizontal velocity component of v1x=1.3 m/s. It's location has been marked at 0.100 s intervals. a.) How long should the ball take to fall on the floor from the time it leaves the edge of the table? b.) How far will the ball travel in horizontal direction from the edge of the table before it hits the floor?

Jul 27, 2018

#### Answer:

a. $t = 0.45 s$
b. $x = 0.59 m$

#### Explanation:

a. Vertical freefall takes place with no regard for any horizontal velocity. The ball will have the same vertical acceleration, and therefore time to the floor, whether it rolls off the table or is just dropped from the height of the table. (Note: this is g, also known as the acceleration due to gravity, doing its thing.)

Horizontally, the initial velocity, $v 1 x = 1.3 \frac{m}{s}$, continues without change while it is falling. The vertical velocity develops without affecting the horizontal velocity.

Therefore it will hit the floor in the same time that it would if simply dropped from the height of the table.

Use the suvat formula

$\textcolor{red}{y = u \cdot t + \left(\frac{1}{2}\right) \cdot a \cdot {t}^{2}}$
where $y = 1 m , u = 0 , a = g = 9.8 \frac{m}{s} ^ 2$ and solve for time.

The term suvat I used above refers to a set of formulas about motion with constant acceleration. This site
http://studywell.com/maths/mechanics/kinematics-objects-motion/suvat-equations/
has 5 formulas in the set. It is often simplified to 4, and I have seen it simplified to 3 formulas. Go to this site if you are not sure what $\textcolor{red}{y = u \cdot t + \left(\frac{1}{2}\right) \cdot a \cdot {t}^{2}}$ is all about.

Plugging the data from our problem into the suggested suvat formula (colored $\textcolor{red}{red}$) and solving for t, we find the answer to part a.
$1 m = \frac{1}{2} \cdot 9.8 \frac{m}{s} ^ 2 \cdot {t}^{2}$

${t}^{2} = \frac{1 \cancel{m}}{\frac{1}{2} \cdot 9.8 \frac{\cancel{m}}{s} ^ 2} = \frac{1}{\frac{4.9}{s} ^ 2} = 0.2041 {s}^{2}$

$t = \sqrt{0.2041 {s}^{2}} = 0.45 s$

b. The horizontal distance ball will travel in time t at a horizontal velocity of 1.3 m/s is calculated as follows. Since

$\text{velocity" = "distance"/"time}$

Using a bit of algebra, we discover how to find $\text{distance}$ when you know $\text{velocity" and "time}$

$\text{distance" = "velocity" * "time}$

$x = {v}_{\text{1x}} \cdot t = 1.3 \frac{m}{\cancel{s}} \cdot 0.45 \cancel{s} = 0.59 m$

I hope this helps,
Steve