Agolfer, standing on a fairway, hits a shot to a green that is elevated 7.67 m above the point where she is standing. if the ball leaves her club with a velocity of 38.0 m/s at an angle of 33.9 â° above the ground, find the time that the ball is in the air before it hits the green.

4.22 seconds First, determine the vertical velocity of the ball. v = 38sin(33.9) = 38 * 0.591666616 = 22.48333142 m/s The altitude the ball will be at time T is given by d = vT - 0.5AT^2 where d = distance v = initial velocity A = acceleration due to gravity T = time. So plug in those values we know d = vT - 0.5AT^2 7.67 m = 22.483 m/s T - 0.5 9.8 m/s^2 T^2 7.67 m = 22.483 m/s T - 4.9 m/s^2 t^2 Convert into a quadratic equation 7.67 m = 22.483 m/s T - 4.9 m/s^2 t^2 0 = 22.483 m/s T - 4.9 m/s^2 T^2 - 7.67 m We now have a quadratic equation with a = -4.9, b = 22.483, and c = -7.67 Solve using the quadratic formula. Getting 2 roots at 0.371172 seconds and 4.217195 seconds. Let's see what we make of those 2 values and see which one answers our question. The golfer hits the ball and 0.371 seconds later the ball has climbed to the same level as the green. The ball continues to climb until its vertical velocity is eventually eliminated and it starts to fall. 4.217 seconds after being hit, the ball finally has fallen on the green. So the answer is 4.217 seconds, which when rounded to 3 significant figures is 4.22 seconds.

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