A football player kicks a ball horizontally off a hill with an initial velocity of 42.0 m/s. It travels a horizontal distance of 59.2m. How y’all was the hill he kicked the ball off of?

9.74 m

Explanation:

Given the following :

Initial Velocity(u) = 42m/s

Horizontal distance (d) = 59.2m

Using the motion equation:

H = ut + 0.5×g×t^2

Where H = height, g = acceleration due to gravity (9.8m/s^2), t = time, u= Initial Velocity

Time(t) = horizontal distance / velocity

t = 59.2 / 42 = 1.41seconds

Therefore,

H = (0 × 1.41) + 0.5(9.8)×(1.41)^2

H = 0 + 9.74169

H = 9.74m

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y