Question

An object is thrown straight downwards at 7 m/s from the top of a water well. If the water we is 72 meters deep, how long, in seconds, does it take the object to hit the bottom of the Well?

Answer 1

Initial velocity u = Tm/s depth of well (h) = 72 m h = ut + 1/2 g t^2 72 = 7 t + 1/2 (9.8) t^2 72 = 7 t + 4.9 t^2 4.9 t^2 + 7t - 72 = 0 t = -7 plusminus squareroot 7^2 - 4 times 4.9 times (-72)/2 times 4.9 t = -7 plusminus 38.21/9.8 time always positive therefore t = -7 = 38.21/9.8 = 3.185 seconds therefore time taken the object to hit the bottom of the well is t = 3.185 s