Answer: action and reaction forces and are the subject of Newton's third law of motion. Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction
Explanation:
1) Data:
Vo = 20 m/s
α = 37°
Yo = 0
Y = 3m
2) Questions: V at Y = 3m and X at Y = 3 m
3) Calculate components of the initial velocity
Vox = Vo * cos(37°) = 15.97 m/s
Voy = Vo * sin(37°) = 12.04 m/s
4) Formulas
Vx = constant = 15.97 m/s
X = Vx * t
Vy = Voy - g*t
Y = Yo + Voy * t - g (t^2) / 2
5) Calculate t when Y = 3m (first time)
Use g ≈ 9.8 m/s^2
3 = 12.04 * t - 4.9 t^2
=> 4.9 t^2 - 12.04t + 3 = 0
Use the quadratic equation to solve the equation
=> t = 0.28 s and t = 2.18s
First time => t = 0.28 s.
6) Calculate Vy when t = 0.28 s
Vy = 12.04 m/s - 9.8 * 0.28s = 9.3 m/s
7) Calculate V:
V = √ [ (Vx)^2 + (Vy)^2 ] = √[ (15.97m/s)^2 + (9.30 m/s)^2 ] = 18.48 m/s
tan(β) = Vy/Vx = 9.30 / 15.97 ≈ 0.582 => β ≈ arctan(0.582) ≈ 30°
Answer: V ≈ 18.5 m/s, with angle ≈ 30°
8) Calculate X at t = 0.28s
X = Vx * t = 15.97 m/s * 0.28s = 4,47m ≈ 4,5m
Answer: X ≈ 4,5 m
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