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I don’t understand how they’re getting the formula of Vx(t) at the start by integrating could you show me how to resolve this in steps please

I don’t understand how they’re getting the formula of Vx(t) at the start by integrating could you show me how to resolve this in steps please

Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter4: The Laws Of Motion
Section: Chapter Questions
Problem 10P
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I don’t understand how they’re getting the formula of Vx(t) at the start by integrating could you show me how to resolve this in steps please
Problem 1-2D Kinematics and drag. * An object having mass m = 1 kg is launched with initial velocity at t=0, v(0) = Vxoi + Vyoj, with Vxo = Vyo = 200 m/s, and initial position x(0) = y(0) = 0. The forces acting of the object are gravity, Fg = -mg j, where you can approximate g = 10 m/s², and drag, Fo= -bv, where b = 0.1 kg/s. (a) Find the time, tmax, it takes for the object to reach maximum height; (b) Find the velocity along the x-axis at maximum height, vx(tmax); (c) Find the value of the maximum height, ymax=y(tmax). (Hint: use In3 = 1.1). (a) The equation of motion (Newton's Il law) V m ,dvx = -bvx ⇒v₂(t)=v₂.e ² at M (mduy = -mg-buy -witten Exam 3 September 2021 Solving for tmox: tmax = To bn (1 + 1/²) tmax = 10n3=11s (b) At t=tmox, v₂ (tmax) = √₂e tmax/20 v₂ (tmax)= V₂o = 67 m/s 1+ yo VT →200 1+200 J00 The solution of y-component is: - VT Vy(t) = (V₁ + Vyo)et/to_ where To = 1/ = 1 = 10 m VT = mg = 1x/0= 100 m/s (c) We can obtain an expression for yit) by integrating d y =Vylt) = (V₁ + Vyo) e t/to_ we obtain: y(t)= -(v +vy)e t/to -v₁t+c where c is an integration constant At t=0, ylo) = C-²₂ (V₁ + Vy₂) = 0 and so c= (v₁+₁)%o and
Transcribed Image Text:Problem 1-2D Kinematics and drag. * An object having mass m = 1 kg is launched with initial velocity at t=0, v(0) = Vxoi + Vyoj, with Vxo = Vyo = 200 m/s, and initial position x(0) = y(0) = 0. The forces acting of the object are gravity, Fg = -mg j, where you can approximate g = 10 m/s², and drag, Fo= -bv, where b = 0.1 kg/s. (a) Find the time, tmax, it takes for the object to reach maximum height; (b) Find the velocity along the x-axis at maximum height, vx(tmax); (c) Find the value of the maximum height, ymax=y(tmax). (Hint: use In3 = 1.1). (a) The equation of motion (Newton's Il law) V m ,dvx = -bvx ⇒v₂(t)=v₂.e ² at M (mduy = -mg-buy -witten Exam 3 September 2021 Solving for tmox: tmax = To bn (1 + 1/²) tmax = 10n3=11s (b) At t=tmox, v₂ (tmax) = √₂e tmax/20 v₂ (tmax)= V₂o = 67 m/s 1+ yo VT →200 1+200 J00 The solution of y-component is: - VT Vy(t) = (V₁ + Vyo)et/to_ where To = 1/ = 1 = 10 m VT = mg = 1x/0= 100 m/s (c) We can obtain an expression for yit) by integrating d y =Vylt) = (V₁ + Vyo) e t/to_ we obtain: y(t)= -(v +vy)e t/to -v₁t+c where c is an integration constant At t=0, ylo) = C-²₂ (V₁ + Vy₂) = 0 and so c= (v₁+₁)%o and
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