Time for rocket shot upward to reach the ground

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SUMMARY

The discussion focuses on solving a physics problem involving a model rocket that releases its first stage at 100.0 m above the ground with an initial upward velocity of 40.0 m/s. The key equation used is Y = Yo + VyoT + 1/2GT^2, where Y is the final position, Yo is the initial position, Vyo is the initial velocity, and G is the acceleration due to gravity. A common mistake identified was neglecting to use a negative value for G, which led to an incorrect calculation resulting in an imaginary number. The correct values are Y = 0, Yo = 100 m, Vyo = 40.0 m/s, and G = -9.8 m/s².

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helpmedude

Homework Statement


A model rocket releases its first stage when it is 100.0 m above the ground and moving upward at 40.0 m/s. Ignoring air resistance, find:

the time it takes the first stage to hit the ground

Homework Equations


Y=Yo+VyoT+1/2GT^2

Quadratic

The Attempt at a Solution



I got to [-(40)+-Squareroot(1600-1960)]/9.8

the problem is that the square root of a negative number is an imaginary number so I am just wondering what I am missing here?
 
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Hi helpmedude,

helpmedude said:

Homework Statement


A model rocket releases its first stage when it is 100.0 m above the ground and moving upward at 40.0 m/s. Ignoring air resistance, find:

the time it takes the first stage to hit the ground

Homework Equations


Y=Yo+VyoT+1/2GT^2

Quadratic

The Attempt at a Solution



I got to [-(40)+-Squareroot(1600-1960)]/9.8

I don't believe that is correct. What numbers did you use for Y, Y0, Vy0, and G?
 
my bad I forgot to make acceleration due to gravity negative that's why it didnt work geez. thanks anyways. Y=0 Yo=100m Vyo=40.0m/s G=-9.8m/s^2 I used 9.8m/s^2
 

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