A catapult launches a rocket at an angle of 46.3° above the horizontal with an initial speed of 81 m/s. The rocket engine immediately starts a burn, and for 2.43 s the rocket moves along its initial line of motion with an acceleration of 26.6 m/s2. Then its engine fails, and the rocket proceeds to move in free-fall.
(a) Find the maximum altitude reached by the rocket.
m
(b) Find its total time of flight.
s
(c) Find its horizontal range.
m
(a) Find the maximum altitude reached by the rocket.
m
(b) Find its total time of flight.
s
(c) Find its horizontal range.
m
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Catapult must be assumed to be of zero height which is not my image of a catapult - this was IMHO an important OMISSION. The height of the catapult should have been stated in a problem as detailed as it is in other specifics.
The unusual start makes the initial Free-Fall of the rocket begin with initial velocity Vo:
Vo = Vi + at {Vi = 81, a = 26.6, t = 2.43} Vo = 81 +(26.6)(2.43) = 146 m/s
initial angle = 46.3°
The distance traveled during engine burn = d = Vi(t) + 1/2at²
d = (81)(2.43) + (0.5)(26.6)(2.43)² = 197 + 78.5 = 276 m
dx = horizontal travel during engine burn = d cos 46.3° = (276)(cos 46.3) = 191 m
dy = vertical travel during engine burn = d sin 46.3° = (276)(sin 46.3) = 200 m
Vox = initial horizontal Free-Fall velocity = Vo cos 46.3 = 146(cos 46.3) = 101 m/s
Voy = initial vertical Free-Fall velocity = Vo sin 46.3 = 146(sin 46.3) = 106 m/s
t = time to max height *above Free-Fall* release point = Voy/g = 106/9.81 =10.8 s
h = max height above Free-Fall release point = 1/2gt² = (0.5)(9.81)(10.8)² = 572 m
at end of Free-Fall rocket falls through a distance = dy
dy = 200 = Voy(T) + 1/2gT²
0 = 4.905T² + 106T - 200
solve for pos root of T:
T = 1.745 s
Answers:
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(a) Maximum Altitude = h + dy = 572 + 200 = 772 m ANS
(b) Total time of flight = 2.43 + 2t + T = 2.43 + (2)(10.8) + 1.745 = 28.2 s ANS
(c) Range = dx + Vox(2t + T) = 191 + (101)(21.6+1.745) = 191 + 2358 = 2549 m ANS
The unusual start makes the initial Free-Fall of the rocket begin with initial velocity Vo:
Vo = Vi + at {Vi = 81, a = 26.6, t = 2.43} Vo = 81 +(26.6)(2.43) = 146 m/s
initial angle = 46.3°
The distance traveled during engine burn = d = Vi(t) + 1/2at²
d = (81)(2.43) + (0.5)(26.6)(2.43)² = 197 + 78.5 = 276 m
dx = horizontal travel during engine burn = d cos 46.3° = (276)(cos 46.3) = 191 m
dy = vertical travel during engine burn = d sin 46.3° = (276)(sin 46.3) = 200 m
Vox = initial horizontal Free-Fall velocity = Vo cos 46.3 = 146(cos 46.3) = 101 m/s
Voy = initial vertical Free-Fall velocity = Vo sin 46.3 = 146(sin 46.3) = 106 m/s
t = time to max height *above Free-Fall* release point = Voy/g = 106/9.81 =10.8 s
h = max height above Free-Fall release point = 1/2gt² = (0.5)(9.81)(10.8)² = 572 m
at end of Free-Fall rocket falls through a distance = dy
dy = 200 = Voy(T) + 1/2gT²
0 = 4.905T² + 106T - 200
solve for pos root of T:
T = 1.745 s
Answers:
---------------
(a) Maximum Altitude = h + dy = 572 + 200 = 772 m ANS
(b) Total time of flight = 2.43 + 2t + T = 2.43 + (2)(10.8) + 1.745 = 28.2 s ANS
(c) Range = dx + Vox(2t + T) = 191 + (101)(21.6+1.745) = 191 + 2358 = 2549 m ANS