Question

A 0.65 kg rock is projected from the edge of the top of a building with an initial velocity of 9.67 m/s at an angle of 46 degrees above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 16.5 m from the base of the building. How tall was the building. Assume the ground is level and the side of the building is vertical. The acceleration of gravity is 9.8 m/s/s.

Answer 1

12.5 m

Step-by-step explanation:

First find the time, then solve for the height.

Given in the x direction:

s = 16.5 m

u = 9.67 cos 46 = 6.717 m/s

a = 0 m/s²

Find: t

s = ut + ½ at²

16.5 = (6.717) t + ½ (0) t²

t = 2.456 s

Given in the y direction:

u = 9.67 sin 46 = 6.956 m/s

a = -9.8 m/s²

t = 2.456 s

Find: s

s = ut + ½ at²

s = (6.956) (2.456) + ½ (-9.8) (2.456)²

s = -12.5 m

The rock falls 12.5 meters, so the building is 12.5 meters tall.