A golfer had to hit his ball up onto a green that was located 3.5 m above the point from which he was hitting the ball. If the ball left his club with a velocity of 22.5 m/s at an angle of 55 deg above the horizontal, how long did it take for the ball to first hit the ground, and how far did it travel during that time? (Ignore the effects of air resistance) Time for the ball to first hit the ground? ________________ Distance the ball travels before hitting the ground? (dH in the figure below) ________________ dH v 22.5 m/s 3.5 m Ball first hits ground here 055° ball Solution here, intial speed , u = 22.5 m/s theta = 55 degree h0 = 3.5 m let the time taken be t h0 = u * sin(theta) * t - 0.5 * g * t^2 3.5 = 22.5 * sin(55) * t - 0.5 * 9.81 * t^2 solving for t t = 3.56 s the horizontal distance , x = u * cos(theta) * t x = 22.5 * cos(55) * 3.56 m x = 45.9 m Time for the ball to first hit the ground is 3.56 s Distance the ball travels before hitting the ground is 45.9 m .