A ball is thrown upward from the roof of a building 100m tall with an initial velocity of 20m/s when will the ball reach a height of 80m The formula is s = sO + vOt + atē/2 where s = h = the height of the ball off the ground in meters sO = hO = the initial height = 100 m vO = the initial velocity = 20 m/s t = the number of seconds a = g = -9.8 m/sē (the acceleration of gravity) ------------------------ Substituting: h = (100) + (20)t + (-9.8)tē/2 h = 100 + 20t - 4.9tē That's the formula for the height h in meters at any time t seconds. Plug in h = 80m and solve for t 80 = 100 + 20t - 4.9tē 4.9tē - 20t - 20 = 0 If you like, clear of decimals by multiplying thru by 10 49tē - 200t - 200 = 0 Use the quadratic formula and you'll get t = -.83 and 4.92, approximately Discard the negative answer. The legitimate answer is 4.92 seconds. Analysis: at 0 seconds, the height is 100 m at 1 second, the height is 115.1 m at 2 seconds the height is 120.4 m then it starts falling at 3 seconds, its height is 115.9 m at 4 seconds its height is 101.6 m at 4.92 seconds its height is 80m at 5 seconds its height is 77.5 m at 6 seconds its height is 43.6 m at 7 seconds its height is 0, i.e., it hits the ground. Edwin