Question

A rocket is launched at an angle of 50.0° above the horizontal with an initial speed of 102 m/s. The rocket moves for 3.00 s along its initial line of motion with an acceleration of 30.0 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.
  m

(b) Find its total time of flight.
  s

(c) Find its horizontal range.

Answer 1

initial speed = Vi = 102 m/s
initial angle = 50° above horizontal
acceleration along 50° line of travel = 30.0 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 = 102t + (0.5)(30.0)t² = 102(3) + 15.0(3)² = 441 m
altitude of d = 441 sin 50 = 337.8 m
range of d = 441 cos 50 = 283.4 m
initial velocity at engine quit = Vi + at = 102 + (30)(3) = 192 m/s
initial vertical velocity at engine quit = 192 sin 50 = 147 m/s
initial horizontal velocity at engine quit = 192 cos 50 = 123.4 m/s

time for rocket to reach max height after engine quit = 147/g = 147/9.81 = 14.98 s
max height of rocket = 337.8 + 147(14.98) - (0.5)(9.81)(14.98)² = 1439.17 m ANS (a)

max velocity of rocket as it falls back to earth = Vf = √2g(1439.17) = 168.03 m/s
avg velocity of rocket as it falls back to earth = 168.03 /2 = 84.015 m/s
time to fall back to earth = 1439.17/84.015 = 17.13 s
total time of flight = 3.00 + 14.98 +17.13 = 35.11 s ANS (b)

range of rocket = 283.4 + 123.4(14.98 +17.13) = 4245.774 m ANS (c)