What is the maximum height to which a stone thrown vertically upwards can rise if, after 1.5 s, its speed is halved?

t = 1.5 s.

g = 10 m / s2.

V = V0 / 2.

hmax -?

According to the law of conservation of total mechanical energy: Ek0 = Ek + En.

En = m * g * h.

Ek0 = m * V0 ^ 2/2.

Ek = m * V ^ 2/2 = m * V0 ^ 2/8.

m * V0 ^ 2/2 = m * g * h + m * V0 ^ 2/8.

V0 ^ 2/2 = g * h + V0 ^ 2/8.

h = V0 ^ 2/2 * g – V0 ^ 2/8 * g = 3 * V0 ^ 2/8 * g.

The body moves with the acceleration of gravity g.

h = V0 * t – g * t ^ 2/2.

3 * V0 ^ 2/8 * g = V0 * t – g * t ^ 2/2.

3 * V0 ^ 2/8 * g – V0 * t + g * t ^ 2/2 = 0.

3 * V0 ^ 2 – V0 * 8 * g * t + 8 * g * g * ^ t2 / 2 = 0.

3 * V0 ^ 2 – 120 * V0 + 900 = 0.

V0 ^ 2 – 40 * V0 + 300 = 0.

D = (-40) ^ 2 – 4 * 1 * 300 = 400.

V01.2 = (40 ± 20) / 2.

V01 = 30 m / s, V02 = 10 m / s.

hmax1 = V01 ^ 2/2 * g.

hmax1 = (30 m / s) ^ 2/2 * 10 m / s2 = 45 m.

hmax2 = V02 ^ 2/2 * g.

hmax2 = (10 m / s) ^ 2/2 * 10 m / s2 = 5 m.

Answer: hmax1 = 45 m, hmax2 = 5 m.