How Far Will the Ball Strike the Opposite Wall?

[roam]

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?





$$v_{ix} = 20 cos 40°$$ = 15.3 m/s
$$v_{iy} = 20 sin 40°$$ = 12.9 m/s

Using $$x = v_{x}t$$ to find time;
50 = 15.3 t => t = 3.2 seconds.

We use $$y = v_{iy}t + \frac{1}{2}a_{y}t^2$$ 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?)


$$12.9 \times 3.27 + \frac{1}{2} \pm 9.8 \times 3.27$$ ?

Thanks.

 

[tiny-tim]

… be consistent …

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

Hi roam!

Either … but you must be consistent!

Decide whether you're going to measure y upward or downward.

If downward, then of course g is +9.8, but v_iy is -12.

If upward, then g is -9.8, but v_iy is +12.

 

[roam]

Yes, thank you, Tiny tim. I had to make sure since I was skeptic about this.

I understand how it goes now, thanks.