A rock is shot vertically upward from the edge of the top of a tall building. The rock reaches its maximum height above the top of the building 1.30 s after being shot. Then, after barely missing the edge of the building as it falls downward, the rock strikes the ground 5.90 s after it was launched.
a) With what upward velocity was the rock shot?
b) What maximum height above the top of the building is reached by the rock?
c) How tall is the building?
Hint given: Constant-acceleration equations, for vertical motion. For the descent from maximum height, you know the initial velocity, the time, and the (downward) acceleration and you want the final velocity at the ground and the maximum height. Next, use the time needed to go from building top to maximum height to get the corresponding distance.