horizontal velocity has to be vo/2, as that is the only component of velocity at the highest peak.
V0^2=Vv^2 + (Vo/2)^2
Vv^2=3Vo^2/4
Vv= sqrt3/2 Vo
launch angle=arctan Vv/Vh= (sqrt3/2 vo)/Vo/2= sqrt3
angle= sixty degrees
A projectile is launched with an initial speed V0. At its highest point, the projectile's speed is V0/2 . What was the launch angle above the horizontal?
(in deg)
1 answer
1 answer