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I was given $\vec{v}(t) = 40 \hat{i} + (30-10t)\hat{j}$ m/s and i found position vector by

$\int \vec{v}(t) = 40\hat{i} + (30-10t)\hat{j}$

which is $\vec{r}(t) = 40\hat{i} + (30t+5t^2)\hat{j} + 50$

Now, I'm supposed to find the greatest height possible, but I don't know how to. How do I find this?

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First, $r(t) = (40t,30t - 5t^2)$ ( Yours is with a $+$ sign). This means: $x = 40t, y = 30t - 5t^2$, can you eliminate $t$ and obtain $y = f(x)$ which is a quadratic function and you can find $y_{max}$.

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First of all

$r(t) = (40t)i+ (30t-5t^2)j + r_0$

Specify $r_0$

If by height it is meant the j-th component of position vector, then simply find t for

$\frac{d(r\cdot j)}{dt} = 0$

Rest is obvious

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  • $\begingroup$ Also problem states that projectile is launched froma 50meter cliff, does that mean $r_0$ is 50? $\endgroup$ – Kibbles Jun 5 '15 at 6:16
  • $\begingroup$ $r_0=0i+50j$ then. $\endgroup$ – Alexey Burdin Jun 5 '15 at 6:50

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