# 2-Dimensional motion problem im stuck

r3dxP
9. A rocket is launched at an angle of 53.0degrees above the horizontal with an initial speed of 100m/s. It moves for 3.00s along its initial line of motion with an acceleration of 30.0 m/s^2. At this time its engines fail and the rocket proceeds to move as a projectile.
A. What is the maximum altitude reached by the rocket?
Ytotal=Y1+Y2
Y1=(1/2)at^2; y=(1/2)(30.0m/s^2)(9.00s^2); y=13.5m
After 3seconds, the initial velocity is 90.0m/s ( 30.0m/s^2 * 3.00s = 90.0m/s ), so we calculate how long it takes until the gravity acting on the velocity cancel each other out, making the velocity=0m/s. Thus, we find the point where it stops decelerating; (90.0m/s) / (9.8m/s^2) = 9.18s.
Y2=velocity_initial*time + (1/2)(acceleration)(time^2);
Y2=(90.0m/s)*sin(53degrees)*(9.18sec)+(-4.9m/s^2)(9.18s)^2 ~= 247meters.
Ytotal=13.5meters+247meters = 260meters

B. What is its total time of flight?
3.00s+9.18s = 12.2sec; since the maximum altitude is 260.m, we use the free-fall equation, squareroot(2y/g) = squareroot[(520.m) / (9.8m/s^2)], which equals 53.1s. Now, we add the time it took until velocity reached the value of 0, and the time it took for the rocket to fall, so we add these two values, 12.2s+53.1s=65.3s

C. What is its horizontal range?
i tried the A part and B part, but i'm not sure if i did it correctly. If i did this wrong, please tell me and help me with it. C, i have no idea how to do, so any help will be nice :)

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C. How does the vertical speed affect the range? How about the horizontal speed? If I shot two rockets perfectly horizontally from a height y at speeds $v_1$ and $v_2$, how long (time) would the flight take in each case?