Free Fall/Constant Acceleration Problem

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An object is launched upwards at 25 m/s, and the problem is to determine when it reaches 20 meters above the ground. The initial variables include an initial height of 0, an initial velocity of 25 m/s, and an acceleration due to gravity of -9.8 m/s². The equation x = V_0 t - (1/2)gt² is used to find the times, which should yield two solutions due to the object's ascent and descent. The proposed answers of 0.994 seconds and 1.557 seconds are discussed, with a suggestion to show the quadratic equation setup for clarity. The discussion emphasizes the importance of solving for both ascent and descent times.
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cheerspens said:

Homework Statement


An object is launched upwards with a velocity of 25 m/s. At what time (or times) will the object be located twenty meters above the ground?


Homework Equations





The Attempt at a Solution


I started off with the following variable list:
Xo=0 Vo=25
X=0 V=0
t=? a=-9.8
Which seems very off to me.
When trying to solve I got 2.55 seconds. But is there another time where it reaches 20m?
How do you solve this?
Thanks!

rock.freak667 said:
x=V_0 t-\frac{1}{2}gt^2


will help you. Remember it is going up, reaching maximum height and falling back down to the ground. So at two times it will be 20m up in the air.

Would my answers be 0.994sec and 1.557sec?
 
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I got one of those answers but not the other!
Maybe you should show your work, in particular the quadratic equation so we can see what your a, b and c numbers are for entry into the quadratic formula.
 

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