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How would I find the height of a building? I have been given the initial velocity, the angle at which the object was thrown, and the landing distance. And I was able to successfully compute the horizontal displacement and the time. I'm just stuck at the height. Any suggestions?

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  • $\begingroup$ Suggestion: look at the equations your book gives, then look at what variables you have, then see how you might be able to rearrange some of them for a different variables and lastly, solve. $\endgroup$ – Kyle Kanos Feb 3 '15 at 2:01
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You just solve for the $t$ (time) corresponding to your $x$ (the landing distance) and evaluate your $y$ (height) by that $t$:

If $x_{(t)}=t\cdot v_0\cdot \cos (\beta )$, then $t_{(x)} = x/v_0\cdot \sec (\beta )$

So you plug that in for $t$ and evaluate the height $y$ by distance $x$:

$y_{(x)}=t_{(x)}\cdot v_0\cdot \sin (\beta )-g\cdot t_{(x)}^2/2$

Also see Wikipedia: derivation of motion

Example:

If $v_0$ = 100 m/s, $\beta$ = 45° and the height you are looking for is $h$ = 200 m you pass this height after 3.86 s and 273.08 m or 10.57 s and 747.33 m.

Solution

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You already have $t_{on\ the\ ground}$ expressed through initial velocity, angle and landing distance. Plug this time into equation for vertical displacement, and you will find the height.

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