Problem 73.

A rock is shot up vertically upward from the edge of the top of the building. The rock reaches its maximum height 2 s after being shot. Them, after barely missing the edge of the building as it falls downward, the rock strikes the ground 8 s after it was launched. Find

(a) upward velocity the rock was shot at;

(b) the maximum height above the building the rock reaches; and

(c) how tall is the building?

Disregard air resistance.

Solution:

We have free fall motion of the rock. The equations, which describe the motion of the rock are the following:

Where – initial velocity, - initial height (height of the building).

(a) At the maximum height the velocity of the rock is zero. Then from the second equation we can find the initial velocity (since t= 2s)

(b) Then we can substitute the initial velocity, the traveled time (2 s) into the first equation and find the maximum height above the building:

(c) We know that after 8 s the rock hits the ground. It means that the position of the rock at this moment of time is 0. Then we substitute this time and this position into the first equation and find the height of the building: