## Statement

From a height of 40 meters an object of negligible mass is thrown downward with a velocity of 20 m/s. How long will it take to hit the ground? What will be its velocity on impact?

## Solution

It's a vertical launch motion. This kind of motion is uniformly accelerated rectilinear motion (u.a.r.m.). The u.a.r.m. equations are as follows:

$$y={y}_{0}+{v}_{0}t+\frac{1}{2}a{t}^{2}$$

$$v={v}_{0}+a\cdot t$$

If we consider the origin of coordinate at the point where the body touches the ground, and the downward direction negative, the data of the problem are as follows:

y_{0}=40m

y=0

v_{0} = -20 m/s

a= -9.81m/s

t?

v?

From the first equation of the u.a.r.m we have...

$y={y}_{0}+{v}_{0}t+\frac{1}{2}a{t}^{2}$ =>

0=40 -20t +(1/2)(-9.81)(t)^{2}

It is a quadratic equation from which we only keep the positive solution t =>

t= 1.47s

That is the time it takes to reach the ground.

To calculate the velocity, we apply the other equation...

$v={v}_{0}+a\cdot {t}^{2}$ =>

v = -20 -9.81(1.47) = -34.42 m/s

Where the sign only indicates the direction (downward) of the motion.

In short, the body takes 1.47 s and reaches a velocity of - 34.42 m/s.

Finally, a clarification. Even though the statement tells us that the mass is negligible, as you can see in the formulas, no magnitude depends on it. Said in a different way, regardless of the mass of the body, in the above-mentioned circumstances, the body will always take 1.47 s to reach the ground and would reach it with the calculated velocity of - 34.42 m/s.

$$\overline{)t=1.47s}\phantom{\rule{0ex}{0ex}}\overline{)v=-34.42m/s}$$