Hi! Can someone help me figure out how to solve this : The height in feet of an object is given by y = -16t2 + 80t + 96, where y is the height and t is the time in seconds that the object is in the air. After how much time does the object hit the ground?
y is the position of the subject, and t is the number of seconds. You need find the time when it hits the ground, i.e. to find the value(s) of t when y=0.
Set \(\large\color{slate}{ 0=-16t^2 + 80t + 96 }\) and solve for t.
should I use the distance formula? meh I'm confused lol
not when you are solving a quadratic equation, no
forget about the words, just set it equal to zero and solve what @SolomonZelman wrote
oh ok I get it thank you! 'll post the answer after I solve it.
HI!!
sure:)
Hi ( :
divide all by \(-16\) first, it will make it loads easier
ah I can't see what you typed I have a "math processing error"
she said that dividing the quadratic equation by -16 would make it a lot easier to solve.(divide every term in the equation by -16)
refresh your browser, usually works
btw, math process error is something that I often have. Switching browsers also works (in fact, more than refreshing, for me).
does -2 or -3 sound right?
as the solution set
no
\[0 = -16 (t-6) (t+1)\]
no
thanks @misty1212 I'm working on it
1/2 or -1/8?
@SolomonZelman when I divided by -16 I got -t^2 + -5t + -6 = 0
solution set of 6,-1
but when I tried to plug it in it didn't work out
\(\large\color{slate}{\displaystyle 0=-16t^2 + 80t + 96}\) \(\large\color{slate}{\displaystyle 0=-16(t^2 -5t -6)}\) \(\large\color{slate}{\displaystyle 0=-16(t+1)(x-6)}\)
for t, in a quadratic equation you DO get -1 and 6, but remember, that you are finding "the seconds' because t is seconds.
And you can't have -1 seconds.
so, you only include the positive solution for t.
True, thank you! :D
yw
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