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.
Best Answer
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).