Answer:
3.7 m
Explanation:
ASSUMING this means extra distance beyond where the cannonball would land WITHOUT the wind assistance but in general ignoring air resistance. Hmmmmmm...tricky
The ball drops from vertical rest to ASSUMED horizontal ground 15 m below in a time of
t = √ (2h/g) = √(2(15)/9.8) = 1.75 s
Without the tail wind, the ball travels horizontally
d = vt = 68(1.75) = 119 m
The tailwind exerts a constant acceleration on the ball of
a = F/m = 12/5.0 = 2.4 m/s²
The average horizontal velocity during the flight is
v(avg) = (68 + (68 + 2.4(1.75)) / 2 = 70.1 m/s
so the distance with tailwind is
d = v(avg)t = 70.1(1.75) = 122.675 m
The extra distance is 122.675 - 119 = 3.675 = 3.7 m
