Posted by **Kaan** on Wednesday, October 17, 2012 at 4:17pm.

A ball is shot from the top of a building with an initial velocity of 19m/s at an angle theta = 45 above the horizontal.

Vxo and Vyo, are 13.4m/s

If a nearby building is the same height and 50m away, how far below the top of the building will the ball strike the nearby building?

I found the time from x=vt, 50=13.4*t

t= 3.73s

then y= yo + Vyo*t - 0.5*at^2

y= yo + 13.4*3.73 - 0.5*9.8*(3.73)^2..

shouldn' t the answer with that part of formula which is " Vyo*t - 0.5*at^2 " give me how far it could be?

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