I need someone who's good at using the quadratic formula
Please post your question.
There @silversurfer
Can you tell under what condition the ball hits the ground ?
h=0
The height has to become zero,right? So, you make h=0 and equate the quadratic part to zero. Then, you just solve for x, in this case, this is the variable t.
The quadratic equation can be rewritten as 5t^2+5t-45=0. Then you can simplify it further by dividing with 5 on both sides, doing which gives you t^2+1-9=0. Now, you can use the quadratic formula given below.
\[t^2 -t-9=0\]
using the quadratic formula: \[t=\frac{-b \pm \sqrt{b^2 -4ac}}{2a}\]
A correction to my previous equation, it would be t^2+t-9=0. Sorry, it was careless of me.
yes indeed: t^2 + t - 9 =0 :P
Then use the formula given by sayak to solve for t. Keep in mind, the equation is of the form ax+bx+c=0 where x is t in this case.Hope this helps.
yes but how do i solve for t and find out the seconds?
\[t=\frac{\sqrt{37}-1}{2}\]
a=1,b=1,c=-9, plug in the values in the formula now. I haven't checked but if you get a negative time as one of the roots, you can rule that out.
which should approximately be equal to 2.5
@getoffmyback got it?
yes I'm following
Clearly speaking, there would be two answers. Check for yourself to see which one is reasonable.
yes it gives me an option of just one
@sayakdbz I have to round to the nearest hundredth
nearest hundredth? hm...
That means one of the answers won't be reasonable. Don't be daunted at that, you will understand for yourself once you calculate the two values.
yea @sayakdbz
I already gave the answer needed: 2.5 which is: t=\frac{sqrt{37}-1}{2}
\[t=\frac{\sqrt{37}-1}{2}\]
rounding 2.5 to the nearest 100th
oh I thought that was the tenth
my bad
wait, sorry :P i approximated the original answer rounded off to the nearest 100th the answer is 2.54 seconds
well thank u all for your help @silversurfer @sayakdbz and @isag
@getoffmyback u are most welcome
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