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The height in meters of a projectile can be modeled by h = −4.9t2 + vt + s where t is the time (in seconds) the object has been in the air, v is the initial velocity (in meters per seconds) and s is the initial height (in meters). A soccer ball is kicked upward from the ground and flies through the air with an initial vertical velocity of 4.9 meters per second. After how many seconds does it land approximately?

Mathematics
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you just confused me.
My bad. Are you familiar with calculus, namely derivatives?
nope.

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Other answers:

OK. No problem. I can help you solve it using another method. Stand by
\[-4.9t^2+49t=0\]solve for \(t\)
h = −4.9t2 + vt + s First let's try to understand this formula and what it means. h is telling you how high the projectile is at ANY point in time. -4.9t^2 is telling you that gravity is acting downward. vt is allowing you to specify the initial velocity (they gave you this information, namely 4.9 meters per second). s is telling you where the projectile is starting (they also told you this, namely at the ground or ZERO). How are you doing so far?
ok. i got that.
satellite73 gave you the short and sweet version. Let me know if you know how to solve a quadratic.
yea i do.
i got t=o,-10
Satellite also gave you the wrong formula. He wrote 49t. That means the ball is moving initially at 49 meters per second. Is that what the problem gave you?
t=0 is a correct solution. That is telling you that the ball was at a height of ZERO, the clock started. Your second solution is the answer you need.
Solve 0=-4.9t^2+4.9t You should get t=0 t=.9574 (not rounded)
ohhh. yay! thanks!

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