# 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.

## 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.

Determine vertical and horizontal velocity compoents.

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

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.
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