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?


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:




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:



v0 = -20 m/s

a= -9.81m/s



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

y=y0+v0t+12at2  =>

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=v0+at2  =>

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.

t=1.47 sv=-34.42 m/s

Formulas worksheet

These are the main formulas that you must know to solve this exercise. If you are not clear about their meaning, we recommend you to check the theory in the corresponding sections. Furthermore, you will find in them, under the Formulas tab, the codes that will allow you to integrate these equations in external programs like Word or Mathematica.