A launch of negligible mass

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:

y₀=40m

y=0

v₀ = -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.

$$\begin{gathered} \boxed{t=1.47 s} \\ \boxed{v=-34.42 m/s} \end{gathered}$$