From a high tower with an interval of τ = 1 s, two stones are thrown with zero initial speed.
| May 13, 2021 | education
From a high tower with an interval of τ = 1 s, two stones are thrown with zero initial speed. At what distance from each other will the stones be at the moment when the speed of the second stone becomes equal to v = 30 m / s?
τ = 1 s.
V01 = V02 = 0 m / s.
V2 = 30 m / s.
g = 10 m / s2.
S12 -?
S12 = S1 – S2.
S1 = V01 * t1 + g * t12 / 2 = g * t12 / 2.
S2 = V02 * t2 + g * t22 / 2 = g * t22 / 2.
t1 = t2 + τ.
V2 = V02 + g * t2 = g * t2.
t2 = V2 / g.
t2 = 30 m / s / 10 m / s2 = 3 s.
t1 = 3 s + 1 s = 4 s.
S1 = 10 m / s2 * (4 s) 2/2 = 80 m.
S2 = 10 m / s2 * (3 s) 2/2 = 45 m.
S12 = 80 m – 45 m = 35 m.
Answer: the distance between the stones will be S12 = 35 m.
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