Find The Velocity Vector Of The Ball 1.85 S After It Is Thrown.?

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

Best Answer

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 tanθ =18.13/15.7
hope u got it

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