Arock 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?

h=23.67 m  : Building height

Explanation:

The rock describes a parabolic path.

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x = xi + vx*t   Equation (1)

Where:  

x: horizontal position in meters (m)

xi: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m/s  

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:

y= y₀+(v₀y)*t - (1/2)*g*t² Equation (2)

vfy= v₀y -gt Equation (3)

Where:  

y: vertical position in meters (m)  

y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Data

v₀ = 12.2 m/s  , at an angle  α=53° above the horizontal

x= 25 m , y=0

Calculation of the time it takes for the ball to hit the ground

We replace data in the equation (1)

x = xi + vx*t

x= 25 m   ,xi=0 ,  vx= v₀*cosα = (12.2 m/s)*cos(53°) =7.34 m/s

25 = 0 + 7.34*t

t= 25 / 7.34

t= 3.406 s

Calculation of the Building height

v₀y =  v₀*sinα = (12.2 m/s)*sin(53°) = 9.74 m/s

in t= 3.406 s, y=0

We replace data in the equation (2)

y= y₀ + (v₀y)*t - (1/2)*gt²

0=  y₀ + (9.74)*(3.406 )- (1/2)*(9.8)(3.406 )²

0=  y₀ + 33.17- -56.84

0=  y₀ - 23.67

y₀ =  23.67 m =h: Building height

it breaks and dirupture other materials and other sencews from the bottom

i'm on my way to history not a mistory.