The ball was thrown vertically upward at a speed of 4 m / s. In 0.8 s, it will cover the path.

At the top of the ball trajectory V = 0:

V = V0 – gt = 0,

where V0 is the initial speed of the ball, V0 = 4 m / s;

g – acceleration due to gravity, g = 10 m / s2.

Ball lifting time: t1 = V0 / g = 4/10 = 0.4 s.

Maximum lifting height of the ball:

h1 = V0 t1 −g t1 ^ 2/2 = 4 × 0.4 – 10 × 0.4 ^ 2/2 = 0.8 m.

Ball drop time: t2 = t – t1 = 0.8 – 0.4 = 0.4 s.

Lowering the ball in time t2:

h2 = g t ^ 2 2/2 = 10 × 0.4 ^ 2/2 = 0.8 m.

Thus, in time t = 0.8 s the ball will cover the path:

s = h1 + h2 = 0.8 + 0.8 = 1.6 m.