Answer:
Explanation:
We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write
where
is the distance of the new object from the sun (orbital radius)
is the orbital period of the object
is the orbital radius of the Earth
is the orbital period the Earth
Solving the equation for , we find
Answer:
B) 27.3 m
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 = vx*t Equation (1)
Where:
x: 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:
(vfy)² = (v₀y)² - 2g(y- y₀) 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 = 30 m/s , at an angle α=40.0° above the horizontal
v₀x = vx = 30*cos40° = 22.98 m/s
v₀y = 30*sin40° = 19.28 m/s
y₀ = 2m
y = 18.0 m
g = 9.8 m/s²
Calculation of the time (t) it takes for the rock to reach at 18 m above the ground
We replace data in the equation (2)
(vfy)² = (v₀y)² - 2g(y- y₀)
(vfy)² = (19.28)² - 2(9.8)(18- 2)
(vfy)² = 371.86 - 313.6
(vfy)² = 58.26
vfy = 7.63 m/s
We replace vfy = 7.63 m/s in the equation (2)
vfy = v₀y - gt
7.63 = 19.28 - (9.8)(t)
(9.8)(t) = 11.65
t = 11.65 / (9.8)
t = 1.19 s
Horizontal distance from where the rock was thrown to the window
We replace t = 1.19 s , in the equation (1)
x = vx*t
x = (22.98)* ( 1.19 )
x = 27.3 m