## Statement

A golf ball rolls with constant velocity on the surface of a table 2.5 m above the ground. When it reaches the edge, it falls off the table as if horizontally launched so that at 0.4 s it is at a horizontal distance of 1 m from the edge of the table. Determine:

a) What was the constant velocity of the ball while rolling on the table?

b) Would you know how to determine how far horizontally will be the ball when it hits the ground?

c) What is the distance of the ball from the ground at 0.4 s?

## Solution

**Data**

Height of the ball. H = 2.5 m

Horizontal distance. x = 1 m at t = 0.4 s.

Acceleration of gravity. g = 9.8 m/s^{2}

**Preliminary considerations**

As described in the statement, the ball behaves as a horizontal launched projectile, so to solve this problem we will need to use the equations for this type of motion.

**Resolution**

a) Knowing that the ball is on x = 1 m at t = 0.4 s, we will just substitute these values into the equation of the horizontal launch to calculate the velocity of the ball in the x-axis. Remember that in this motion the body moves with uniform rectilinear motion:

$$x={x}_{0}+v\xb7t\Rightarrow \phantom{\rule{0ex}{0ex}}1m=0m+v\xb70.4s\Rightarrow \phantom{\rule{0ex}{0ex}}v=\frac{1m}{0.4s}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)v=2.5m/s}$$

b) To determine how far the ball will be when it hits the ground we must determine when the impact will occur. To do this, first we determine how long it takes using the equation for position on the y-axis and make y = 0, that is, the ball is on the ground:

$$y=H-\frac{1}{2}\xb79.8m/{s}^{2}\xb7{t}^{2}\Rightarrow \phantom{\rule{0ex}{0ex}}0m=2.5m-\frac{1}{2}\xb79.8m/{s}^{2}\xb7{t}^{2}\Rightarrow \phantom{\rule{0ex}{0ex}}{t}^{2}=\frac{-2.5m}{-4.9m/{s}^{2}}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)t=0.71s}$$

Now we only have to substitute the time calculated in the equation of position for the x-axis:

$$x={x}_{0}+v\xb7t\Rightarrow \phantom{\rule{0ex}{0ex}}x=0m+2.5m/s\xb70.71s\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)x=1.77m}$$

c) Finally, we can determine the height above the ground the ball is located at t = 0.4 s, using of the equation of position for y for this time:

$$y=H-\frac{1}{2}\xb7g\xb7{t}^{2}\Rightarrow \phantom{\rule{0ex}{0ex}}y=2.5m-\frac{1}{2}\xb79.8m/{s}^{2}\xb70.{4}^{2}\Rightarrow \phantom{\rule{0ex}{0ex}}\overline{)y=1.72m}$$