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Toranc3
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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