Answer:
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
This question involves the concept of the equations of motion in the vertical motion as well as the horizontal motion.
The building is "23.6 m" tall.
First, we consider the horizontal motion of rock. Assuming the friction to be negligible, the horizontal motion must be uniform. Thus the following equation shall be used for the horizontal motion:
[tex]s=v_xt[/tex]
where,
s = horizontal range = 25 m
[tex]v_x[/tex] = horizontal component of initial speed = vCosθ = (12.2 m/s)Cos53°
t = time taken for the motion = ?
Therefore,
[tex]t=\frac{25\ m}{(12.2\ m/s)(Cos\ 53^o)}[/tex]
t = 3.4 s
Now, we consider the vertical motion. Applying the second equation of motion:
[tex](h-h_o)=v_{iy}t+\frac{1}{2}gt^2[/tex]
where,
h = final height = 0 m
h₀ = height of building = ?
[tex]v_{iy}[/tex] = vertical component of initial speed = (12.2 m/s)Sin 53° = 9.74 m/s
g= -9.81 m/s²
Therefore,
[tex]0\m-h_o=(9.74\ m/s)(3.4\ s)+\frac{1}{2}(-9.81\ m/s^2)(3.4\ s)^2\\\\h_o=56.7\ m-33.1\ m\\[/tex]
h₀ = 23.6 m
Learn more about equations of motion here:
brainly.com/question/20594939?referrer=searchResults
The attached picture shows the equations of motion.
