Agolfer takes a swing from a hill twenty feet above the cup with an initial upwards velocity of 32 feet per second. how long does it take the ball to land on the ground near the cup?

2.5 seconds The acceleration due to gravity is about 32 ft/s^2. So formula for distance under constant acceleration and with an initial velocity is: d = VT - 0.5 AT^2 where V = initial velocity T = time A = acceleration due to gravity. Substituting known values and solving for T gives -20 = 32 ft/s * T - 0.5 * 32 ft/s^2 * T^2 -20 = 32 ft/s * T - 16 ft/s^2 * T^2 0 = 32* T - 16 * T^2 + 20 -16T^2 +32T +20 = 0 And we now have a quadratic equation with A=-16, B = 32, C = 20. Use the quadratic formula to find the roots, which are -0.5 and 2.5. The -0.5 root would be what happened if you managed to go back in time 0.5 seconds, dig underground 20 feet and hit the ball. But we'll ignore that and use the 2.5 time instead. So the ball lands on the ground 2.5 seconds after being hit.