Question

The problem goes like this: If you throw an object upward, and you know that it reaches 50% of it's maximum height after 2s, what is the maximum height.

So we know: Acceleration : -9.8m/s^2
and time:2s = 0.5y

I can't figure out how to solve this

Answer 1

The problem goes like this: If you throw an object upward, and you know that it reaches 50% of its maximum height after 2 s, what is the maximum height.

So we know: Acceleration: $-9.8\frac{m}{s^2}$ and time: $2s = 0.5y$

Maximum height equals: $\frac{v^2}{2} = gh$

Time needed to reach 50% of maximum height: $\frac{h}{2} = vt - \frac{gt^2}{2}$

So, we have system of equations:

$$\left\{ \begin{array}{c} \frac{v^2}{2} = gh \\ \frac{h}{2} = vt - \frac{gt^2}{2} \end{array} \right.$$

Or:

$$\left\{ \begin{array}{c} v = \sqrt{2gh} \\ h = 2vt - gt^2 \end{array} \right.$$

Substitute (1) equation to second:

$$h - 2\sqrt{2gt}\sqrt{h} + gt^2 = 0$$

We have quadratic equation with roots:

$$\sqrt{h} = \frac{2\sqrt{2gt} \pm \sqrt{(2\sqrt{2gt})^2 - 4gt^2}}{2} = \frac{\sqrt{2gt} \pm \sqrt{gt^2}}{1} = \sqrt{gt}(\sqrt{2} \pm 1)$$

Or for height:

$$\begin{array}{l} h = gt^2(3 \pm 2\sqrt{2}) = 9.8\frac{m}{s^2} * 4s^2(3 \pm 2\sqrt{2}) \\ h_1 = 228\,m \\ h_2 = 6.7\,m \\ \end{array}$$

Answer: 228 or 6.7 m