Actually there is only one solution and that is 2.9s because when u substitute the values into the quadratic eq. your 12.5 & 4.9 should be negative but what you did is all positive, that is why you have two solutions for this problem. Remember: when the object reach the maximum high, the velocity will become zero and the time needed to reach maximum high from the ground will be equal to the time that object needs to return to its initial position (ground).,,thanks a lot for your work,,
haha i came from Marsh too!!!!!!
JFMfilms 1 week ago
thumbs up if you came here from marsh!
fotobik 1 month ago 3
yeah man i get it,,,thanks a lot,
galachoxy 3 months ago
Actually there is only one solution and that is 2.9s because when u substitute the values into the quadratic eq. your 12.5 & 4.9 should be negative but what you did is all positive, that is why you have two solutions for this problem. Remember: when the object reach the maximum high, the velocity will become zero and the time needed to reach maximum high from the ground will be equal to the time that object needs to return to its initial position (ground).,,thanks a lot for your work,,
galachoxy 3 months ago
@galachoxy I think the quad equation is just fine.. check it out again
steelback999 3 months ago
shouldn't 1/2 become negative?
Zadose18 1 year ago
@Zadose18 yes...y does 1/2 did ot become negative?
ibabaonon 1 year ago
this is was really very helpful for me to understand !! thank you !!
bluebird8526 1 year ago
i think you made a mistake in adding 12.5+6.22 =18.7, you put 1.8 instead... just letting you know..
bluebird8526 1 year ago