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 
:
If you put 7% into a decimal it would be 0.07
Answer:
32 I would say is the correct answer to this question
Answer:
Equation of tangent plane to given parametric equation is:
Step-by-step explanation:
Given equation
---(1)
Normal vector tangent to plane is:
Normal vector tangent to plane is given by:
![r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]](https://tex.z-dn.net/?f=r_%7Bu%7D%20%5Ctimes%20r_%7Bv%7D%20%3Ddet%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Chat%7Bi%7D%26%5Chat%7Bj%7D%26%5Chat%7Bk%7D%5C%5Ccos%28v%29%26sin%28v%29%260%5C%5C-usin%28v%29%26ucos%28v%29%261%5Cend%7Barray%7D%5Cright%5D)
Expanding with first row
at u=5, v =π/3
---(2)
at u=5, v =π/3 (1) becomes,
From above eq coordinates of r₀ can be found as:
From (2) coordinates of normal vector can be found as
Equation of tangent line can be found as:
Answer:
-6r+6
Step-by-step explanation:
Let us expand the expression:
8-(6r+2)=
8-6r-2=
6-6r
OR
-6r+6
<em>I hope this helps! :)</em>
