Please help me out!....

On an alien planet with no atmosphere, acceleration due to gravity is given by g = 12m/s^2. A cannonball is launched from the or

almond37 [142]

Answer:

a) $\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$ , b) $\theta = \frac{\pi}{4}$ , c) $y_{max} = 84.375\,m$ , $t = 3.75\,s$ .

Step-by-step explanation:

a) The function in terms of time and the inital angle measured from the horizontal is:

$\vec r (t) = [(v_{o}\cdot \cos \theta)\cdot t]\cdot i + \left[(v_{o}\cdot \sin \theta)\cdot t -\frac{1}{2}\cdot g \cdot t^{2} \right]\cdot j$

The particular expression for the cannonball is:

$\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$

b) The components of the position of the cannonball before hitting the ground is:

$x = (90\cdot \cos \theta)\cdot t$

$0 = 90\cdot \sin \theta - 6\cdot t$

After a quick substitution and some algebraic and trigonometric handling, the following expression is found:

$0 = 90\cdot \sin \theta - 6\cdot \left(\frac{x}{90\cdot \cos \theta} \right)$

$0 = 8100\cdot \sin \theta \cdot \cos \theta - 6\cdot x$

$0 = 4050\cdot \sin 2\theta - 6\cdot x$

$6\cdot x = 4050\cdot \sin 2\theta$

$x = 675\cdot \sin 2\theta$

The angle for a maximum horizontal distance is determined by deriving the function, equalizing the resulting formula to zero and finding the angle:

$\frac{dx}{d\theta} = 1350\cdot \cos 2\theta$

$1350\cdot \cos 2\theta = 0$

$\cos 2\theta = 0$

$2\theta = \frac{\pi}{2}$

$\theta = \frac{\pi}{4}$

Now, it is required to demonstrate that critical point leads to a maximum. The second derivative is:

$\frac{d^{2}x}{d\theta^{2}} = -2700\cdot \sin 2\theta$

$\frac{d^{2}x}{d\theta^{2}} = -2700$

Which demonstrates the existence of the maximum associated with the critical point found before.

c) The equation for the vertical component of position is:

$y = 45\cdot t - 6\cdot t^{2}$

The maximum height can be found by deriving the previous expression, which is equalized to zero and critical values are found afterwards:

$\frac{dy}{dt} = 45 - 12\cdot t$

$45-12\cdot t = 0$

$t = \frac{45}{12}$

$t = 3.75\,s$

Now, the second derivative is used to check if such solution leads to a maximum:

$\frac{d^{2}y}{dt^{2}} = -12$

Which demonstrates the assumption.

The maximum height reached by the cannonball is:

$y_{max} = 45\cdot (3.75\,s)-6\cdot (3.75\,s)^{2}$

$y_{max} = 84.375\,m$

7 0

3 years ago

6 [ 13 - 2 ( 4 + 1 ) ] simply step by step

sp2606 [1]

Answer:

18

Step-by-step explanation:

$6(13 - 2(4 + 1)) \\ 6(13 - 2 \times 5) \\ 6(13 - 10) \\ 6 \times 3 \\ = 18$

4 0

3 years ago

Read 2 more answers

Below are two parallel lines with a third line intersecting them.

Allushta [10]

180-48= 132 ( I think)

8 0

3 years ago

Best answer get Brainliest. 50 points.

Oliga [24]

<span>The equation would be 5x + 0.10y = 35, and since y equals 100, the 0.10y = 10. Then you get your equation which would be 5x + 10 = 35. and you can use subtraction property of equality and subtract 10 from both sides which gives you 25. So x would equal 5. In this case x represents number of dogs walked. Hope this helps! :D</span>

3 0

3 years ago

Read 2 more answers

AB:BC is 3:4. Solve for x.

Sphinxa [80]

I got B but I'll show you how I got that answer.

Let's look at AB and BC. AB = 3 and BC = 4.
3/4 = 0.75

On the triangle, AB = 90. And BC = 10x - 20. We need to solve for x and prove that AB/BC = 0.75.

Let's start with Choice A.

10(11) - 20 = 90.

AB = 90
BC = 90.

90/90 = 1. It doesn't equal 0.75. Meaning this answer is wrong.

Now let's look at Choice B.

10(14) - 20 = 120.

AB = 90
BC = 120

90/120 = 0.75

So, 3:4 = 90:120 making Choice B the correct Answer. Let me know if this is correct

6 0

3 years ago