Due to gravity, the rock strikes the ground at a horizontal distance of 18.1 m from the base of the building. building 18.1 m 59 ◦ 11.9 m/s h how tall is the building? assume the ground is level and that the side of the building is vertical. the acceleration of gravity is 9.8 m/s 2 . answer in units of m.

The building is 12.56 m tall

Explanation:

Horizontal velocity = 11.9 x cos 59 = 6.13 m/s

Vertical velocity = 11.9 x sin 59 = 10.2 m/s

Horizontal distance traveled = 18.1 m

Consider the horizontal motion of rock,

     We have equation of motion, s = ut + 0.5at²

     Initial velocity, u = 6.13 m/s

     Acceleration, a = 0 m/s²

     Displacement, s = 18.1 m

     Substituting,

                   s = ut + 0.5at²  

                   18.1 = 6.13 x t + 0.5 x 0 x t²  

                       t = 2.95 s

Consider the vertical motion of rock,

     We have equation of motion, s = ut + 0.5at²

     Initial velocity, u = 10.2 m/s

     Acceleration, a = -9.81 m/s²

     Displacement, s = Negative of height = -h

     Time, t = 2.95 s

     Substituting,

                   s = ut + 0.5at²  

                   -h = 10.2 x 2.95 + 0.5 x (-9.81) x 2.95²  

                       h = 12.56 m

     The building is 12.56 m tall

would be false hope this out

v 154.9= 0.9 length length s little over 154.9cm