How Long is the Ball in the Air Using a Quadratic Equation?

[caprija]

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?

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

 

[cristo]

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

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

This is not the quadratic formula.

 

[caprija]

I know it's not the quadratic formula, I was just asking. I figured out the height, it's 21.2 and the answer to the question is 4 sec. but I don't know how to get it.

 

[cristo]

For this problem, we don't need to know the actual height. The equation you state gives the displacement of the ball. Now, if we throw a ball from a certain height d into the air, and catch it when it falls back down to the height d, then what is the displacement? [hint: displacement is a vector quantity]

 

[Tedjn]

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 $$h_{\text{original}}=1.2+20t_{\text{original}}-5t_{\text{original}}^2$$. The final position is given by $$h_{\text{final}}=1.2+20t_{\text{final}}-5t_{\text{final}}^2$$. 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.

 

[HallsofIvy]

I know it's not the quadratic formula, I was just asking. I figured out the height, it's 21.2 and the answer to the question is 4 sec. but I don't know how to get it.

WHAT were you asking? You said

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

What exactly was your question?

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?

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.