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How do I figure out the time? quadratic function? -b/2a?

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- Thread starter caprija
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How do I figure out the time? quadratic function? -b/2a?

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cristo

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If the ball is caught at the same height at which it is thrown, what can you say about h?

By the way, I don't know what this is

This is not the quadratic formula.

By the way, I don't know what this is

How do I figure out the time? quadratic function? -b/2a?

This is not the quadratic formula.

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cristo

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To help clarify for the original poster, because I do not know his or her level of mathematical study,

By itself, the formula provided does give how high the ball is above the ground at a certain time*t*. It should be easy, then, to solve for the time that it takes to reach that height; just plug in the right value for *h*.

Except we are not given*h*! No, we are not given *h* explicitly, but it can be figured out easily if you plug in the right value for *t*.

This is analogous to cristo's comment about the displacement. If you take physics (or maybe you have already), the displacement is [final position - original position]. The original position is given by [tex]h_{\text{original}}=1.2+20t_{\text{original}}-5t_{\text{original}}^2[/tex]. The final position is given by [tex]h_{\text{final}}=1.2+20t_{\text{final}}-5t_{\text{final}}^2[/tex]. What are you looking for and how can you simplify?

If you need to, ponder this: why is the height a quadratic equation with two time solutions?

As for -b/2a, that will give the x-coordinate of the vertex. Since the parabola is pointing downwards on a plot of height versus time, it will give the time for the maximum height, which some problems ask for, but not this one.

By itself, the formula provided does give how high the ball is above the ground at a certain time

Except we are not given

This is analogous to cristo's comment about the displacement. If you take physics (or maybe you have already), the displacement is [final position - original position]. The original position is given by [tex]h_{\text{original}}=1.2+20t_{\text{original}}-5t_{\text{original}}^2[/tex]. The final position is given by [tex]h_{\text{final}}=1.2+20t_{\text{final}}-5t_{\text{final}}^2[/tex]. What are you looking for and how can you simplify?

If you need to, ponder this: why is the height a quadratic equation with two time solutions?

As for -b/2a, that will give the x-coordinate of the vertex. Since the parabola is pointing downwards on a plot of height versus time, it will give the time for the maximum height, which some problems ask for, but not this one.

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HallsofIvy

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WHAT were you asking? You said

What exactly was your question?How do I figure out the time? quadratic function? -b/2a?

At what height was it hit- what is h when t= 0? If it was caught at that same height, set h= to that height and solve. Since this is a quadratic equation, it will have two solutions. One is obvious, the other is your answer.caprija said:The height h of the ball is given be h = 1.2 + 20t -5t^2, where t is in seconds. If the ball is caught at the same height at which it was hit, how long is it in the air?

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