Question

While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 6.71 m/s. the stone subsequently falls to the ground, which is 18.9 m below the point where the stone leaves your hand. at what speed does the stone impact the ground? how much time is the stone in the air? ignore air resistance and take g = 9.81 m/s2. (this is not a suggestion to carry out such an experiment!)?

Answer 1

First the stone will move upward till reach maximum height, when it will start to go down till hit the ground. So to calculate the time going up, we can use equation: Vf = Vi + a*t , where Vf is final speed, Vi is initial speed, a is acceleration and t is time In this case a = -g = -9.81 m/s^2 because the stone goes against gravity attraction. 0 = 6.71 - 9.81*t => t = 6.71/9.81 = 0.68 seconds And the distance upward will be: d = Vi*t + 1/2*a*t^2 = 6.71*0.68 + 1/2*(-9.81)*0.68^2 d = 4.59-2.29 = 2.29 metres Then the stone starts to go down till the ground, which is 18.9 + 2.29 metres below and it takes, now a = g = 9.81 m/s2 because stone goes in the same direction than gravity atraction: d = Vi*t + 1/2*a*t^2 => 0*t+1/2*9.81*t^2 => 21.19 = 4.905*t^2 => t = 2.08 seconds So then the stone is on air 0.68 seconds going up and 2.08 going down, that is in total: 2.76 seconds