A ball was thrown with an initial velocity of 25 m/s at an angle of 30 degrees above the horizontal. Find: a.) The time the ball was in the air. b.) The range of the flight of the ball.
can we split the velocity up into its component vectors that match the angle?
at 30 degree we have an up vector of 12.5 and an out vector of 12.5 sqrt(3) if thats right, then the equation for time in the air is: -4.9t^2 +12.5t + h0 :height when thrown"; solving for the zeroes at an initial height of 0 gives us a time of 125/49 seconds in the air the out vector * time in air would amount to range then ... (12.5*sqrt(3)*125)/49 = range but this only counts if my initial assumption is correct :)
amistre64's answer is correct if the height thrown is the same height of landing. This comes about from the integration constants in deriving the equation for time in the air from the fact that y''(t) = - g (gravitational constant)