The body, falling without initial velocity, had a velocity of 20 m / s at an altitude of 60 m from the Earth.

| June 18, 2021 | education

The body, falling without initial velocity, had a velocity of 20 m / s at an altitude of 60 m from the Earth. The total fall height H and the distance h traveled in the last second of the fall are …

h1 = 60 m.
V (60) = 20 m / s.
Vo = 0 m / s.
g = 10 m / s ^ 2.
H -?
h -?
The body moves with constant acceleration g. Let’s find the speed of the body V when hitting the ground.
h1 = (V ^ 2 – V (60) ^ 2) / 2 * g.
V = √ (V (60) ^ 2 + 2 * g * h1).
V = √ (20 m / s) ^ 2 + 2 * 10 m / s ^ 2 * 60 m) = 40 m / s.
Find the height of the fall H = (V ^ 2 – Vо ^ 2) / 2 * g, since Vо = 0 m / s, then H = V ^ 2/2 * g.
H = ((40 m / s) ^ 2/2 * 10 m / s ^ 2 = 80 m.
Let’s find the time of the entire fall: t = V / g.
t = 40 m / s / g = 10 m / s ^ 2 = 4 s.
Let’s find the distance h (3) covered by the body in 3 seconds: h (3) = g * t (3) ^ 2/2.
h (3) = 10 m / s ^ 2 * (3 s) ^ 2/2 = 45 m.
This means that in 4 (fourth) second the body has passed the distance h = H – h (3).
h = 80 m -45 m = 35 m.
Answer: H = 80 m, h = 35 m.



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