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An arrow is fired from a height of 65 m above the ground at an angle of 60 degrees above the horizontal. It hits the ground 210 m away from its firing point. Calculate its initial and final velocity, flight time and maximum height.

What I've done so far: Set the origin at the starting point of the arrow, down is the negative-$y$ direction.

$\Delta x = 210 $ m

$\Delta y = -65$ m

So according to the projectile motion formulas:

$v_y^2 = u_y^2 + 2(-9.8)(-65)$

$-65 = u_yt + 0.5(-9.8)t^2$

$210 = u_xt$

I dont know where to go from here

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closed as off-topic by heather, fffred, user36790, John Rennie, Kyle Kanos Dec 24 '16 at 12:23

This question appears to be off-topic. The users who voted to close gave this specific reason:

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We know:

  1. $\Delta y = v_{iy}\Delta t - \frac{1}{2}g\Delta t^{2} = v_{i} \sin (\theta) \Delta t - \frac{1}{2}g\Delta t^{2}$ (where $g = 9.8 \ m/s^2$)

  2. $v_{ix} = v_{fx} = \frac{\Delta x}{\Delta t} = v_{i} \cos(\theta)$ because $a_{x} = 0 \ m/s^{2}$.

In both cases, $\Delta t$ is the same, so we can substitute $\Delta t = \frac{\Delta x}{v_{i} \cos(\theta)}$ in the first equation, yielding

$\Delta y = \tan(\theta) \Delta x - \frac{1}{2} g \frac{\Delta x^{2}}{v_{i} \cos^{2}(\theta)}$

You know everything but $v_{i}$, so you can do the rest. As for $v_{f}$, you will know $v_{fx}$ from your first calculation and you can afterwards find the final angle; doing some trigonometry, you can find $v_{f}$. Remember, $v_{i}$ and $v_{f}$ are the magnitudes of $\overrightarrow{v_{i}}$ and $\overrightarrow{v_{f}}$ respectively.

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solve what you have to get expresion of t for the vertical component and t for the horizontal component These equations must be equal. Without t in the equation can you now rearrange to solve it for some other parameter - which one?

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