A catapult launches a rocket at an angle of 50.9° above the horizontal with an i

Question

A catapult launches a rocket at an angle of 50.9° above the horizontal with an initial speed of 101 m/s. The rocket engine immediately starts a burn, and for 3.31 s the rocket moves along its initial line of motion with an acceleration of 35.5 m/s2. Then its engine fails, and the rocket proceeds to move in free-fall.

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

Your response differs from the correct answer by more than 10%. Double check your calculations.
(b) Find its total time of flight.

Your response differs from the correct answer by more than 10%. Double check your calculations.
(c) Find its horizontal range.

Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.

Explanation / Answer

here,

initial speed = Vi = 101 m/s

initial angle = 50.9 degree above horizontal

acceleration along 50.4 line of travel = 35.5 m/s^2

time of acceleration along initial line of travel = 3.31 s

d = distance along initial line of travel = Vi(t) + 1/2at^2

d = 101 * t + (0.5)(35.5)t^2 = 97(3.31) + 17.75 * (3.31)^2

d = 515.54 m

altitude of d = 515.54 * sin(50.9) = 400 m

range of d = 515.54 * cos(50.9) = 325 m

initial velocity at engine quit = Vi + at = 101 + (35.5)(3.31) = 218 m/s

initial vertical velocity at engine quit = 218 * sin(50.9) = 169.6 m/s

initial horizontal velocity at engine quit = 218 cos(50.9) = 137.5 m/s

time for rocket to reach max height after engine quit = 169.6/g = 169.6/9.81 = 17.3 s

max height of rocket = 345 + 169.6 * (17.3) - (0.5)(9.81)(17.3)^2 = 1811 m

b)

max velocity of rocket as it falls back to earth = Vf = sqrt(2g(1811)) = 188.5 m/s

avg velocity of rocket as it falls back to earth = 188.5/2 = 94.2 m/s

time to fall back to earth = 1811/94.2 = 19.2 s

total time of flight = 3.31 + 17.3 + 19.2 = 39.8 s

c)

range of rocket = 325 + 137.5 (17.3 + 19.2) = 5343 m

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