The pebble was thrown vertically upward with an initial speed of 20 m / s.

| July 5, 2021 | education

The pebble was thrown vertically upward with an initial speed of 20 m / s. At what height is the speed of the pebble halved? Do not take air resistance into account.

V0 = 20 m / s.

g = 10 m / s2.

V = V0 / 2.

h -?

1) 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.

h = 3 * (20 m / s) ^ 2/8 * 10 m / s2 = 15 m.

2) The body moves with the acceleration of gravity g.

h = (V0 ^ 2 – V ^ 2) / 2 * g = (V0 ^ 2 – V0 ^ 2/4) / 2 * g = 3 * V0 ^ 2/4 * 2 * g = 3 * V0 ^ 2 / 8 * g.

h = 3 * (20 m / s) ^ 2/8 * 10 m / s2 = 15 m.

Answer: at a height of h = 15 m, the speed of the pebble will decrease 2 times.



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