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

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SUMMARY

The discussion focuses on solving the free fall problem for an object launched upwards at a velocity of 25 m/s, determining the times it reaches a height of 20 meters. The equation used is \( x = V_0 t - \frac{1}{2} g t^2 \), where \( g \) is the acceleration due to gravity, set at -9.8 m/s². The calculated times for the object to reach 20 meters are 0.994 seconds and 1.557 seconds, confirming that the object reaches this height twice during its trajectory.

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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!

[tex]x=V_0 t-\frac{1}{2}gt^2[/tex]


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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