Question

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 98 m/s. The rocket moves for 3.00 s along its initial line of motion with an acceleration of 31.0 m/s2. At this time, its engines fail and the rocket proceeds to move as a projectile.

(a) Find the maximum altitude reached by the rocket.

(b) Find its total time of flight.

(c) Find its horizontal range.

Answer 1

initial speed = Vi = 98 m/s
initial angle = 53° above horizontal
acceleration along 53° line of travel = 31.0 m/s²
time of acceleration along initial line of travel = 3.00 s

d = distance along initial line of travel = Vi(t) + 1/2at²
d = 98t + (0.5)(31.0)t² = 98(3) + 15.5(3)² = 433.5 m
altitude of d = 433.5 sin 53 = 346.2 m
range of d = 433.5 cos 53 = 260.8 m
initial velocity at engine quit = Vi + at = 98 + (31)(3) = 191 m/s
initial vertical velocity at engine quit = 191 sin 53 = 152.5 m/s
initial horizontal velocity at engine quit = 191 cos 53 = 114.9 m/s

time for rocket to reach max height after engine quit = 152.5/g = 152.5/9.81 = 15.5 s
max height of rocket = 346.2 + 152.5(15.5) - (0.5)(9.81)(15.5)² = 1531.5 m ANS (a)

max velocity of rocket as it falls back to earth = Vf = √2g(1531.5​) = 173.2 m/s
avg velocity of rocket as it falls back to earth = 173.2 /2 = 86.6 m/s
time to fall back to earth = 1531.5​/ 86.6 = 17.68 s
total time of flight = 3.00 + 15.5 +17.68 = 36.18 s ANS (b)

range of rocket = 260.8 + 114.9(15.5 +17.68) = 4073.18 m ANS (c)