A golfer standing on a fairway, hits a shot to a green that iselevated 5.5m above where she is standing. If the ball leaves herclub with a velocity of 46 m/s at an angle of 35 °above the ground, find the time that the ball is in the air beforeit hits the green.
In part of the listed answer to this problem thereis:
y=Viny*t + .5ayt2
-5.5m= 26.4m/s * t + .5 (-9.8m/s) *t 2
t= 26+-√((262) -4*49(-5.5))
My question is where does the 49 come from? .5 * -9.8 is4.9


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