Question

When a rocket reaches an altitude of 40 m it begins to travel along the parabolic path (y- 40) 160 x, where the coordinates are measured in meters. If the component of velocity in the vertical direction is constant at vy 180 m/s, determine the magnitudes of the rocket's velocity and acceleration when it reaches an altitude of 80 m. 40)2 160x 40 m

Answer 1

V_y = 180 mu/s (y - 40)^2 = 160x differentiate both side w r t 't' d/dt(y - 40)^2 = d/dt(160)x 2(y - 40)(dy/dt) = 160(dx/dt) - (1) at y = 80 m 2(80 - 40) (180)(160)(dx/dt) V_x = 90 r/s v = squareroot V_x ^2 + V_y ^2 = squareroot (90)^2 + (180)^2 201.25 m/s eq (1) on 2(y - 40) Vy = (160)V_x As V_y is constant dbv y/dt = d \r\n (y - 40) v_y = 80 vx differentiate both side w r t 't' d/dt[Vy(y - 40)] = 80d/dt Vx Rightarrow (vy) (dy/dt) = 80 a_x a_x = (180)^2/80 a_x = 405 m/s^2