Problem A stone is thrown from the top of a building with an initial velocity of 24.2 m/s straight upward, at an initial height of 54.1 m above the ground. The stone just misses the edge of the roof on its way down, as shown in Figure 2.20. (a) Determine the time needed for the stone to reach its maximum height. (b) Determine the maximum height. (c) Determine the time needed for the stone to return to the height from which it was thrown, and the velocity of the stone at that instant. (d) Determine the time needed for the stone to reach the ground. (e) Determine the velocity and position of the stone at t = 6.14 s. Strategy The diagram in Figure 2.20 establishes a coordinate system with Yo = 0 at the level of release from the thrower's hand, with y positive upward. Write the velocity and position kinematic equations for the stone and substitute the given information. All the answers come from these two equations by using simple algebra or just substituting the time. For part (a), for example, the stone comes to rest for an instant at maximum height, so set v=0 at this point and solve for time. Then substitute the time into the displacement equation, obtaining the maximum height. Figure 2.20 A freely falling object is thrown upward from a building.