(adsbygoogle = window.adsbygoogle || []).push({}); A ball is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m/s, 40° ABOVE HORIZONTAL. How far above or below its original level will the ball strike the opposite wall?

[tex]v_{ix} = 20 cos 40°[/tex] = 15.3 m/s

[tex]v_{iy} = 20 sin 40°[/tex] = 12.9 m/s

Using [tex]x = v_{x}t[/tex] to find time;

50 = 15.3 t => t = 3.2 seconds.

We use [tex]y = v_{iy}t + \frac{1}{2}a_{y}t^2[/tex] to find How far above or below its original level will the ball strike the opposite wall;

Now I don't know wether I should take acceleration due to gravity, a = 9.8 as positive or negative for this. (What about v_{iy}= 12.0?)

[tex]12.9 \times 3.27 + \frac{1}{2} \pm 9.8 \times 3.27[/tex] ?

Thanks.

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# Homework Help: ± acceleration due to gravity?

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