The body is thrown at an initial speed of 23 m / s at an angle of 400 to the horizon. Neglecting the air resistance, determine for the moment of time 1.1 s after the start of the movement: the speed of the body, the height of the body’s rise.
V0 = 23 m / s.
∠α = 40 °.
t = 1.1 s.
g = 9.8 m / s2.
The motion of a body can be divided into two types: horizontally, it moves uniformly with a speed V0х, vertically uniformly accelerated with an acceleration of gravity g and an initial speed V0у.
V0х = V0 * cosα.
V0у = V0 * sinα.
The height of the rise h is expressed by the formula: h = V0у * t – g * t ^ 2/2 = V0 * sinα * t – g * t ^ 2/2.
h = 23 m / s * sin40 ° * 1.1 s – 9.8 m / s2 * (1.1 s) ^ 2/2 = 10.3 m.
V = √ (Vx ^ 2 + Vy ^ 2).
Vx = V0x = V0 * cosα.
Vх = 23 m / s * cos40 ° = 17.47 m / s.
Vу = V0у – g * t = V0 * sinα – g * t.
Vу = 23 m / s * sin40 ° – 9.8 m / s2 * 1.1 s = 3.94 m / s.
V = √ (17.47 m / s) ^ 2 + (3.94 m / s) ^ 2) = 17.9 m / s.
Answer: the body will move at a speed of V = 17.9 m / s to a height of h = 10.3 m.
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