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A child tosses a ball directly upward. Its total time in the air is T. Its maximum height is H. What is its height after it has been in the air a time T/4? Neglect air resistance.

Ok so I know that there is no x-component to the velocity:

$$D=v_0 t+\frac{1}{2}at^2$$ $$\frac{1}{2}D=H$$ so, $$\frac{1}{2}H=v_0(\frac{1}{4}t)+[(4.9)(\frac{1}{4}t)^2]$$

I am confused as to where to go from here?

I am overcomplicating this, missing the underlying concept. I don't want the answer, but just some guidance.

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First, the problem is not presented clear enough - maybe it was not defined clear enough in the first place. Idea is that you know maximal height (e.g. $H = 4$m) and you have to obtain height for $T/4$. You know neither $T$, nor $v_0$, so you must express them as functions of $H$.

Second, the expression you wrote $$h=v_0 t+\frac{1}{2}at^2$$ is for the displacement (also height in your case) and not for total distance traveled.

Simply put $h = 0$ for $t = T$ and $h = H$ for $t = \frac{T}{2}$ in your equation to obtain expressions for $v_0$ and $T$ as function of $H$ and then you can calculate $h(\frac{T}{4})$...

Ah, and don't forget that $a = -9.8$m/s$^2$ (minus sign).

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  • $\begingroup$ Ok, I have an embarrassing lack of understanding of some of the concepts here. I completely understand that $h=0 for t=T and h=H for t=(T/2)$, but solving for the unknowns is the area I lack understanding. Obviously, I have to go back and re-learn some important math concepts, which I believe will be solving systems of equations. Becoming competent solving Kinematics and Dynamics questions is really important to me, and I am thankful for the guidance. Can anyone recommend a source of practice questions I could learn from? $\endgroup$
    – Kurt
    May 9, 2012 at 6:28
  • $\begingroup$ @Kurt OK, I also realized (after some additional brainstorming), that the problem is also that you must know what are you looking for $h(\frac{T}{4})$ and what real data you have (only $H$). The problem specified is not trivial at all, and neither is the answer. $\endgroup$
    – Pygmalion
    May 9, 2012 at 6:50
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You've gotten very turned around here, and I'm not sure where 1/2D = H came from. Let's start from scratch. I'll provide some leading questions, you provide the answers, and hopefully you'll guide yourself to a solution.

  1. Is there any symmetry you can use in the problem to simplify things a bit? Alternatively, is there any way to consider the problem so that you can set terms you don't know anything about to zero? (These are related questions).

  2. At what time t does the ball reach its maximum height (in terms of T)?

  3. Can you write an equation relating the maximum height H to the time T?

  4. What changes (or doesn't change) when you look at time T/4?

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  • $\begingroup$ 1. The max height of the ball will occur when the distance is at half of its total value (not displacement, but total distance travelled). Thus, 1/2D=H (the max height occurs at half the total distance travelled). $\endgroup$
    – Kurt
    May 9, 2012 at 5:26
  • $\begingroup$ 1. The total displacement will be zero, so I can make D=0. 2. The ball reaches its max height at 1/2T. 3. I want to include acceleration in any equation I choose. (how do you insert line breaks and created bulleted lists? (I tried enclosing my text in <ul> tags, and inserting a </br> to no effect? $\endgroup$
    – Kurt
    May 9, 2012 at 5:41
  • $\begingroup$ @Kurt Formula you are using is for displacement not for distance traveled. You must therefore think in terms of displacement. $\endgroup$
    – Pygmalion
    May 9, 2012 at 6:05

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