A rocket is launched at an angle of 59.0 Solution may b hlpfull.. By v = u + at

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may b hlpfull.. By v = u + at =>v = 75 + 25 x 25 = 700 m/s [initial velocity for 2nd phase at an angle of 53* from horizontal] By s = ut + 1/2at^2 =>s = 75 x 25 + 1/2 x 25 x (25)^2 =>s = 9687.5m Thus the vertical height (h1) = 9687.5 x sin53* = 7736.78m Thus the horizontal distance covered (R1) = 9687.5 x cos53* = 5830.08m For 2nd stage:- (a) the vertical component of initial velocity uy = 700 x sin53 = 559.04 m/s By v^2 = u^2 - 2gs =>0 = (559.04)^2 - 2 x 9.8 x h2 =>h2 = 15945.19m =>the rocket's maximum altitude(H) = h1+h2 = 7736.78 + 15945.19 = 23681.97m (b) Let the rocket take t2 sec to reach h2 =>By v = u - gt =>0 = 559.04 - 9.8 x t2 =>t2 = 57.04 sec Let the rocket take t3 sec to fall H meter:- =>By s = ut + 1/2gt^2 =>23681.97 = 0 + 1/2 x 9.8 x t3^2 =>t3 = sqrt[4833.06] = 69.52 sec Thus total time of the flight (t) = t1 + t2 + t3 = 25 + 57.04 + 69.52 = 151.56 sec (c)The horizontal component of velocity ux = 700 x cos53* = 421.27 m/s By distance = velocity x time =>R2 = 421.27 x (t2+t3) = 421.27 x 126.56 = 53315.93m Thus total R = R1 + R2 = 5830.08 + 53315.93 = 59146.01m

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