I have the answers to this question but I\'m not sure how they got them. The ans
ID: 1333675 • Letter: I
Question
I have the answers to this question but I'm not sure how they got them.
The answers are (a) 1.52 Km, (b) 36.1s, (c) 4.05 Km
A catapult launches a rocket at an angle of 53.0 degrees above the horizontal with an initial speed of 100 m/s. The rocket engine immediately starts a burn, and for 3.00 s the rocket moves along its initial line of motion with an acceleration of 30.0 m/s^2. Then its engine fails, and the rocket proceeds to move in free fall. Find (a) the maximum altitude reached by the rocket, (b) its total time of flight, and (c) its horizontal range.
Explanation / Answer
V = u + at
V = 100 +30*3 = 190 m/s in the line of intial motion
S = 100*3 + 1/2*30*(3)^2 = 435
Sx = 435cos(53) = 261 m
Sy = 435sin(53) = 343.6m
now it start projectile motion after that
H = U^2 sin^2(p)/2g = ((190)^2*(0.79)^2)/(2*9.8) = 1149.49
=> maximum altitude reached by the rocket = 343.6 + 1149.49 = 1.52Km
b)
0-343.6 = 190sin(53)*t -1/2*9.8*t^2
0-343.6 = 150.1*t -1/2*9.8*t^2
=> t = 32.77 sec
total time = 32.77+3 = 35.77 sec
c)
X = 190cos(53)*t = 3735.78
total X = 3735.78 + 261 = 3996.78 m Or 4km