The body is thrown horizontally at a speed of 49 m / s. What are the horizontal and vertical components of the speed of the stone in 0.3 s? After how long will the speed of the stone be directed at an angle of 45 degrees to the horizon?
Data: Vg (horizontal throw speed) = 49 m / s; t (considered fall time) = 0.3 s; α (the required angle of inclination of the vector of the full velocity of the stone to the horizon) = 45º.
Constants: g (acceleration due to gravity) ≈ 9.8 m / s2.
1) a) Horizontal speed in 0.3 s: Vg = const = 49 m / s;
b) Vertical speed after 0.3 s: Vw = g * t = 9.8 * 0.3 = 2.94 m / s.
2) The required duration of the fall: tgα = Vw / Vg = g * t / Vg, whence we express: t = Vg * tgα / g = 49 * tg 45º / 9.8 = 5 s.
Answer: In 0.3 s, the horizontal and vertical speeds will be respectively 49 m / s and 2.94 m / s; full speed will be at 45º to the horizontal after 5 seconds of fall.
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