**A rock is projected from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of 53? above the horizontal. The rock strikes the ground a horizontal distance of 25 m from the base of the building. Assume that the ground is level and that the side of the building is vertical. How tall is the building.**

**The Answer is**

Vertical velocity = 12.2sin53°

Vertical velocity = 9.74m/s

Horizontal velocity = 12.2cos53

horizontal velocity = 7.34m/s

The ball traveled 25m horizontally

time of travel = 25/7.34

Time of travel = 3.40 seconds

The displacement of the ball is the height of the building

The displacement of the ball is the height of the building

Take up direction as positive

s = ut + 0.5*a*t²

s = 9.74*3.40 + 0.5*(-9.8)*3.4²

s = 33.16 - 56.64

s = -23.5m

The displacement calculates correctly as -ve because it is down.

Height of building = 23.5 m.