A model rocket launches with an initial velocity of 7 m/s, at an angle of 81 deg

Question

A model rocket launches with an initial velocity of 7 m/s, at an angle of 81 degrees (with respect to the horizon). The rocket's engine provides an acceleration of 2.3 m/s^2, acting in the same direction as the initial velocity. After 18 seconds, the rocket engine cuts out. a.) What is the maximum height attained by the rocket? ? miles b.) How far does the rocket travel in the horizontal direction? ? feet c.)What is the total flight time? ? minutes

Explanation / Answer

By v = u + at =>v = 7 + 2.3 x 18 = 48.4 m/s By s = ut + 1/2at^2 =>s = 7 x 18 + 1/2 x 2.3 x (18)^2 =>s = 498.6 m Thus the vertical height (h1) = 498.6 x sin81 = 492.46m Thus the horizontal distance covered (R1) = 498.6 x cos81 = 78m For 2nd stage:- (a) the vertical component of initial velocity uy = 48.4 x sin81 = 47.8 m/s By v^2 = u^2 - 2gs =>0 = (47.8)^2 - 2 x 9.8 x h2 =>h2 = 116.59m =>the rocket's maximum altitude(H) = h1+h2 = 492.46 + 116.59 = 609.05 m (c) Let the rocket take t2 sec to reach h2 =>By v = u - gt =>0 = 47.8 - 9.8 x t2 =>t2 = 4.87 sec Let the rocket take t3 sec to fall H meter:- =>By s = ut + 1/2gt^2 =>609.05 = 0 + 1/2 x 9.8 x t3^2 =>t3 = 11.14 sec Thus total time of the flight (t) = t1 + t2 + t3 = 18 + 4.87 + 11.14 = 34 sec (c)The horizontal component of velocity ux =48.4 x cos81* = 7.57 m/s By distance = velocity x time =>R2 = 7.57 x (t2+t3) = 7.57 x 16.01 = 121.19 m Thus total R = R1 + R2 = 78 + 121.19 =199.19m

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