Solve Free Fall Problem: Object Launched at 25 m/s, Reaches 20m at 2.55 s

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An object is launched upwards at 25 m/s and reaches 20 meters above ground at 2.55 seconds. The equation of motion used is x = V_0 t - (1/2)gt^2, where g is the acceleration due to gravity. The discussion reveals that the object will reach 20 meters at two distinct times: once while ascending and once while descending. The calculated times for these instances are approximately 0.994 seconds and 1.557 seconds. Understanding the object's trajectory is crucial for solving similar free fall problems.
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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!
 
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cheerspens said:

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!

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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