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Phoenix [80]
3 years ago
14

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

igin with an initial speed of 90m/s at an angle θ with the horizontal (0 ≤ θ < π/2).
(a) Find a function r(t) that gives the cannonball's position after t seconds. Your expression should be in terms of θ.
(b) Find the initial angle θ for the cannonball's velocity that will maximize the horizontal distance the cannonball will travel before hitting the ground.
(c) If θ = 30°, what is the maximum height of the cannonball? When does it reach that height?
Mathematics
1 answer:
almond37 [142]3 years ago
7 0

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

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Step-by-step explanation:

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In 7-th grade we have:

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Then, the ratio for the 8-th graders must be about the same:

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Ok, we now see that with the bigger option we obtained almost the same ratio (if we use the smaller values for X, we will get a ratio bigger than 0.15, so 0.15 is the better aproximation that we can find to the 0.14 of the 7-th graders)

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<u>A </u><u>pair of shoes</u><u> gives us</u><u> 48 </u><u>different outfits.</u>

What is a linear equation?

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We will employ the basic counting principle, which states that if one task can be completed in method A and another in way B, then both tasks may be completed in way A*B.

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By basic counting principle, number of outfits = A*B*C = 6*2*4 = 48 outfits possible

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Learn more about linear equation

brainly.com/question/11897796

#SPJ4

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