Free Fall/Constant Acceleration Problem

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SUMMARY

The discussion centers on solving a free fall and constant acceleration problem involving an object launched upwards with an initial velocity of 25 m/s. The key equation used is x = V_0 t - (1/2)gt², where g is the acceleration due to gravity, approximately -9.8 m/s². The object reaches a height of 20 meters at two distinct times: approximately 0.994 seconds and 1.557 seconds after launch. Participants emphasize the importance of using the quadratic formula to find both times accurately.

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  • Understanding of kinematic equations in physics
  • Familiarity with the quadratic formula
  • Knowledge of gravitational acceleration (g = -9.8 m/s²)
  • Basic algebra skills for solving equations
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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:
[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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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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