A 5.0 kg cannonball is fired horizontally at 68 m/s from a 15-m-high cliff. A strong tailwind exerts a constant 12 N horizontal force in the direction the cannonball is traveling. What is the extra distance (d)?

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. H...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

unlike poles attract