The ball is thrown vertically upward at a speed V0 = 10 m / s. At what height h will the speed of the ball be half V0?

Ignoring the air resistance, we find that the movement of the ball up to the highest point of the trajectory is uniformly slowed down, with an acceleration g = 9.8 m / s2 directed vertically downward.

Speed equation:

V = V0 – gt,

where V0 is the initial velocity of the ball, V0 = 10 m / s.

At height h, according to the problem statement: V = V0 / 2 = 10/2 = 5 m / s.

Ascent time to height h: t = (V0 – V) / g = (10 – 5) / 9.8 = 0.51 s.

Desired height h:

h = V0 t −g t ^ 2/2 = 10 × 0.51 – 9.8 × 0.51 ^ 2/2 = 3.83 m.

Answer: 3.83 m.