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 :)

 

[Päällikkö]

A. What happened to the 100m/s initial speed?
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?

 

[quicknote]

To calculate the horizontal range, we can use the equation R=Vx*t, where R is the range, Vx is the horizontal component of the initial velocity, and t is the total time of flight. In this case, Vx=100m/s*cos(53degrees) = 58.3m/s. Plugging in the values, we get R=(58.3m/s)*(65.3s) = 3800m. Therefore, the horizontal range of the rocket is approximately 3800 meters.

As for parts A and B, your calculations seem to be correct. To confirm, you can also use the equation Vf=Vi+at to calculate the final velocity after the rocket's engines fail, and then use that value to calculate the maximum height reached using the equation Vf^2=Vi^2+2ay. Both methods should give you the same answer of approximately 260 meters. Additionally, you can check your answer for the total time of flight by using the equation t=(Vf-Vi)/a to calculate the time it takes for the rocket to reach its maximum height, and then adding that to the time it takes for the rocket to fall back to the ground, which should also give you a total of approximately 65.3 seconds. Overall, your approach to solving the problem was correct.