Physics, 16.11.2019 00:31 AnNguyen

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?

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Answer from: maddysmall32

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

Arock is projected from the edge of the top of a building with an initial velocity of 12.2 m/s at an

Answer from: Quest

B) as the temperature increases, the kinetic energy of the particles increases.|-/

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we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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