From a balcony at a height of 20 m above the ground, a ball was thrown into the courtyard with an initial speed of 8 m / s, directed horizontally. What coordinates will the ball have in 2 s. after throwing?
h0 = 20 m.
V0 = 8 m / s.
g = 10 m / s2.
t = 2 s.
The ball will move horizontally evenly, so its coordinate will change according to the law: x = V0 * t.
x = 8 m / s * 2 s = 16 m.
The ball will move vertically downward at uniform acceleration with the acceleration of gravity g, the initial vertical component of the velocity equal to 0.
h = h0 – g * t ^ 2/2.
h = 20 m – 10 m / s2 * (2 s) ^ 2/2 = 0.
Answer: the ball will have coordinates x = 16 m, h = 0. It will be on the surface of the earth at a distance of x = 16 m from the balcony.
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