# 1D Motion under Gravity

I have attempted this question in so many different ways but I am getting nowhere. Could someone point me in the correct direction so I can work it out myself?

Here is the question:

An object $K$ is shot upwards with a velocity $v_i$, after $t$ seconds it is 50 metres high and time $2t$ seconds it is 80 metres high. Solve for $v_i$ and $t$.

I have tried breaking it into two parts but I always hit a wall.

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Can you show how far you got? – Bernhard Jul 28 '13 at 13:45

Sometimes a good picture is worth a hundred words.

If you can take it from here, don't read the rest of the answer.

So the system of equations will be:

$$\left( v_i - \frac{g\tau}{2}\right)\tau=50 \\ \left( v_i - \frac{3g\tau}{2}\right)\tau=30$$

Subtracting them we get:

$$g\tau^2=20$$

Assuming $g \approx 10$, we will have:

$$\Rightarrow \tau = \sqrt {2} \\ v_i = \frac{60}{\sqrt 2}$$

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Thank you sir, that is a perfect explanation. Once I saw the diagram it all became clear – user Jul 28 '13 at 14:10
A small question, would you mind telling me how you got 3gt in equation 2. – user Jul 28 '13 at 15:10
Area of a trapezoid is $\frac{a+b}{2}h$. – Ali Jul 28 '13 at 15:18
I got it thanks!! – user Jul 28 '13 at 15:34