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A girl is throwing a ball.

If I have an equation $h= -4.9*t^2+20t+1.2$, where $t$ is number of second after the girl has thrown the ball, and $h$ is the height above ground. I want to find out when will the ball hit the ground. My solution is set $-4.9*t^2+20t+1.2=0$, and then I get $t=4.1$ seconds. Is it correct?

An it is wrong to say that we solve $-4.9*t^2+20t+1.2=4.1$ and then solve for $t$. Am I right?

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    Looks fine, about $4.14$, though that kind of precision is kind of silly, since the model is a poor one, unless the girl is on an airless planet. This is by setting $h=0$. The calculation mentioned at the end has no relevance to the problem.2012-11-04
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    @André Nicolas You means solving -4.9*t^2+20t+1.2=4.1 for t, makes no sense. AM I right?2012-11-04
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    Yes, sorry, "the calculation mentioned at the end" was perhaps ambiguous. Setting $h=4.1$ makes no sense at all.2012-11-04
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    @André Nicolas Thanks2012-11-04

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So that the question will not remain unanswered, here is an answer.

You calculated the (positive) value of $t$ such that $-4.9t^2+20t+1.2=0$ correctly, and this was the right thing to do.

I get that $t\approx 4.14$. That kind of precision (to the nearest hundredth of a second), is physically unreasonable, since the quadratic model assumes that there is no air resistance.

As to the final question, setting $-4.9t^2+20t+1.2=4.1$ has no connection to solving the problem of finding the time when the ball hits the ground.

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    The units are metres on both sides. Solving $-4.9t^2 + 20t + 1.2 = 4.1$ gives the time when the ball is $4.1$ metres above the ground. It doesn't solve the problem posed but it is a valid thing to do.2012-11-04