A small stone is projected vertically upwards from a point O with a speed of 19.6m/s. Modelling the stone as a particle moving freely under gravity, find the length of time for which the stone is more than 14.7 m above O

We want to find out the time during which the stone is above a certain point. As this stone is thrown upwards we know that the first time it passes this point it will be on its way up and when it passes the point the second time it will be falling. Therefore we know that the difference between these times is the amount of time that the stone spent above this point. Now we know that we are trying to find out the times at which the stone is at this point we can gather information needed for the SUVAT equations. We know that the acceleration is due to gravity and so it is 9.8m/s. But as the gravity is opposing the initial velocity of the stone we make this value negative, -9.8 m/s. We know the initial speed is 19.6 m/s and that S is 14.7m above O. We now have three values so will be able to work out the fourth. We want a suvat equation that involves Time, Acceleration, Distance and Initial velocity. So we take the equation s = ut + 1/2(at^2). With our example 14.7 = 19.6t - 1/2(9.8t^2), if we rearrange this formula we get a typical quadratic formula. 4.9t^2 - 19.6t + 14.7 we can work out both values of t from this. t = 3 and t = 1. The difference between these times is 2 seconds hence the stone was above this height for 2 seconds.