The ball was thrown upward at a speed of 10 m / s. At what height will the ball’s speed be 2 times less than the initial one?
Data: V1 (speed of throwing the ball up) = 10 m / s; V2 (speed at the desired altitude) = 0.5V1.
Constants: g (acceleration due to gravity) ≈ 10 m / s2.
We express the desired height at which the ball will have to be from the equality: Ek1 = Ek2 + En2.
m * V1 ^ 2/2 = m * V2 ^ 2/2 + m * g * h.
V1 ^ 2/2 = (0.5V1) ^ 2/2 + g * h.
V1 ^ 2/2 = V1 ^ 2/8 + g * h.
h = (V1 ^ 2/2 – V1 ^ 2/8) / g.
h = (10 ^ 2/2 – 10 ^ 2/8) / 10 = 3.75 m.
Answer: The speed of the ball in question will be 2 times less than the initial one at a height of 3.75 m.
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