Answer:
Step-by-step explanation:
The vertical component of the initial launch can be found using basic trigonometry. In any right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse. Let the vertical component at launch be 
. The magnitude of 
will be the hypotenuse, and the relevant angle is the angle to the horizontal at launch. Since we're given that the angle of elevation is 
, we have:
(Recall that 
)
Now that we've found the vertical component of the velocity and launch, we can use kinematics equation 
to solve this problem, where 
is final and initial velocity, respectively, 
is acceleration, and 
is distance travelled. The only acceleration is acceleration due to gravity, which is approximately 
. However, since the projectile is moving up and gravity is pulling down, acceleration should be negative, represent the direction of the acceleration.
What we know:
Solving for 
:
Answer:
.324 or 18.9 degrees
Step-by-step explanation:
rule is
sine equals opposite over hypotenuse,
cosine equals adjacent over hypotenuse, and
tangent equals opposite over adjacent
OR
soh cah toa
sine = soh = 12/37 = .324
arcsin(.324) or sin^-1(.324) in DEGREES not radians =
18.9 degrees
Answer:
2.62
Step-by-step explanation:
First, write the square root as exponent.
Move the denominator to the numerator and negate the exponent.
Use log product property.
Use log exponent property.
Substitute values.
Radius, r = 3
The equation of a sphere entered at the origin in cartesian coordinates is
x^2 + y^2 + z^2 = r^2
That in spherical coordinates is:
x = rcos(theta)*sin(phi)
y= r sin(theta)*sin(phi)
z = rcos(phi)
where you can make u = r cos(phi) to obtain the parametrical equations
x = √[r^2 - u^2] cos(theta)
y = √[r^2 - u^2] sin (theta)
z = u
where theta goes from 0 to 2π and u goes from -r to r.
In our case r = 3, so the parametrical equations are:
Answer:
x = √[9 - u^2] cos(theta)
y = √[9 - u^2] sin (theta)
z = u
From lowest to highest
22.075, 23.015, 25.085, 25.125
