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## Homework Statement

a rocket is launched at an angle of 53 degrees above the horizontal with an initial speed of 100 m/s. the rocket moves for 3 s along its initial straight line of motion with an acceleration of 30m/s^2. FInd the max altitude reached by the rocket

## Homework Equations

y=y0 + v0*t +1/2*g*t^(2)

v0y=v0sin(53)

v0x=v0cos(53)

vy=v0y+g*t

## The Attempt at a Solution

velocity components:

v0y=100m/s*sin(53)=79.86m/s

v0x=100m/s*cos(53)=60.18m/s

Found the vertical distance with the acceleration of 30m/s^(2) for 3sec

y=y0 + v0*t +1/2*g*t^(2)

y=79.86m/s*3 sec + 1/2*30m/s*(3sec)^(2) =374.58m

now I am trying to find my final velocity for the acceleration of 30m/s^(2) for 3sec because that will be my initial velocity for the acceleration of 9.8m/s^(2)

I use this formula, vy=v0y+gt

I am not sure which v0y do I use. Do i just use the initial velocity of 100m/s or do I use my v0y that I calculated?(100*sin(53)=79.86m/s) Which one do I use and why? thanks