Q. A man on the edge of a cliff H = 40 m high throws a ball directly upward. It returns past him 1.8 s later. (H actually is the height of the point of release of the ball above the base of the cliff. Neglect air resistance.) (Hint - Gravity causes a downward acceleration at the rate g = 9.81 m/s2.) a) With what initial speed did the man throw the ball? Ans. H = 40m t = 1.8 seconds Using, x-x0 = v0t + 1/2 at^2 To find vo: x -xo = 0 ( I wonder why x -xo , is zero,) when the height of the cliff is given to be 40m. 0 = 1.8u + 4.9 * (1.8)^2 Solving for u, we get: v = 8.82 m/s, which is correct I think! b) How high above the point of release did the ball go? We can use, v^2 = v0^2 + 2g (x-xo) v = final velocity = 0 m/s v0 from part a, is 8.82 m/s and g = 9.81 m/s2 Plugging in all that, we get: x-x0 = 3.964 m c) What is the speed of the ball when it passes him on the way back down? Same as a) 8.82 m/s d) How long after the man throws it does the ball hit the ground at the bottom of the cliff? This is where I am stuck...... Which equation to use? v^2 = u^2 + 2g ( x - xo) or x - xo = v0t + 1/2 gt^2 and what would my variables be v0 = 8.82 m/s g = 9.81 m/s2 We need to find 't' and add it to the 1.7seconds in the problem That would be the total time, to hit the ground, Plz. help!!!