A ball is dropped from a height. If it requires 0.200 s to cross the final 6.00 m before striking the ground, determine the height, in m, from which it was dropped. Take g = 10 m/s ^2 .

Step-by-step solution

Let the velocity of the ball at the beginning of the last 6.00 m be u . Using the equation of motion for the last 6.00 m : s = ut + 1 2 gt^2 6.00 = u(0.200) + 1 2 (10)(0.200)^2 6.00 = 0.200u + 5(0.04) 6.00 = 0.200u + 0.20 0.200u = 5.80 u = 29 m/s This velocity is acquired by the ball after falling a height h₁ from rest. Using v^2 = u^2 + 2gh₁ where initial velocity is zero: (29)^2 = 0 + 2(10)h₁ 841 = 20h₁ h₁ = 42.05 m The total height from which the ball was dropped is H = h₁ + 6.00 . H = 42.05 + 6.00 = 48.05 m