Question

A rocket starts from the rest 100[m] above the ground and is launched such that it accelerates at a 30degree incline from the horizontal at a rate of 6.78[m/s^2]. If the wind resistance along the horizontal direction is 2.87[m/s^2], find the maximum elevation, range and flight time of the rocket.

Answer 1

Given that :

initial velocity of rocket, v₀ = 0 m/s

distance from ground to rocket, y = 100 m

horizontal acceleration, a_x = 6.78 m/s²

using equation of motion 3,

v² = v₀² + 2 a y                                                          { eq.1 }

inserting the values in above eq.

v² = 2 (6.78 m/s²) (100 m)

v = 1356 m²/s²

v = 36.8 m/s

The final vertical velocity is given by, v_fy = v_f sin                                                { eq.2 }

where, = inclination angle = 30 degree

inserting the values in eq.2,

v_fy = (36.8 m/s) sin 30⁰

v_fy = 18.4 m/s

time taken by the rocket is given by,   v_fy = v₀ + a_x t                                                 { eq.3 }

inserting the values in eq.3,

(18.4 m/s) = (0 m/s) + (2.87 m/s²) t

t = 6.41 sec

Now, the vertical displacement during this time is given by :

y = [(v_fy + v₀) / 2] t                                                               { eq.4 }

inserting the values in eq.4,

y = [(18.4 m/s) + (0 m/s) / 2] (6.41 s)

y = 58.9 m

(a) The maximum elevation is given as :

H = y + y                                                                          { eq.5 }

inserting the values in eq.5,

H = (100 m) + (58.9 m)

H = 158.9 m

(c) The time of flight is given as :

H = v_0y + (1/2) a t²

(158.9 m) = (0 m/s) + (0.5) (2.87 m/s²) t²

t² = 111.1 s²

t = 10.5 sec