A football is thrown horizontally with an initial velocity of (15.7 m/s)x^. Ignoring air resistance, the average acceleration of the football over any period of time is (-9.81 m/s2)y^.

(a) Find the velocity vector of the ball 1.85 s after it is thrown.

v = (_____ m/s)x^ - (______ m/s)y^

(b) Find the magnitude and direction of the velocity at this time.

Magnitude: ________m/s

Direction: _____ degrees below horizontal

I cannot for the life of me figure out this question. Any help would be appreciated! Thank you in advance!

*Posted: May 19, 2013*

A)as there is no acceleration in horizontal direction horizontal velocity remains same(ie. 15.7m/s) and as there is no initial velocity in vertical direction but acceleration of -9.8

by v=u+at

as u=0

a=-9.8

t=1.85s

therefore v=1.85*(-9.8)=-18.13

therefore the vector is 15.7x^ -18.13y^

magn.=[(15.7)^2 -(18.13)^2]^1/2

for the direction tanÎ¸ =18.13/15.7

hope u got it