physics
posted by Anonymous .
A projectile is launched at a 60 degree angle above the horizontal with an initial velocity of 35.5 m/s. What is the projectile's velocity (magnitude and direction) 1.50 s into the flight?

Vo = 35.5m/s @ 60o.
Xo = 35.5*cos60 = 17.75 m/s.
Yo = 35.5*sin60 = 30.74 m/s.
Y = Yo + g*t
Y = 30.74  9.8*1.5 = 16.04 m/s.
tanA = Y/Xo = 16.04/17.75 = 0.90366
A = 42.1o
V=Xo/cosA=17.75/cos42.1=23.9 m/s.@42.1o.
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