Aball is shot straight up from the surface of the earth with an initial speed of 19.6 m/s. neglect any effects due to air resistance. how much time elapses between the throwing of the ball and its return to the original launch point?

4 s

Explanation:

u = 19.6 m/s, g = 9.8 m /s^2

Let the time taken to reach the maximum height is t.

Use first equation of motion.

v = u + at

At maximum height, final velocity v is zero.

0 = 19.6 - 9.8 x t

t = 19.6 / 9.8 = 2 s

As the air resistance be negligible, is time taken to reach the ground is also 2 sec.

So, total time taken be the ball to reach at original point = 2 + 2 = 4 s