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

a) , b) , c) , .

Step-by-step explanation:

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

The particular expression for the cannonball is:

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

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

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

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

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

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

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

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

Which demonstrates the assumption.

The maximum height reached by the cannonball is:

5x+40

5(x + 8)

7x - 49

7(x - 7)

18x + 9

9(2x + 1)

15x - 10

5(3x - 2)

step-by-step explanation:

for number one all you have to do is factor 5 out of 5x + 40

for number 2 factor 7 out of 7x - 49

for number three factor 9 out of 18x + 9

for number four factor 5 out of 15x - 10

step-by-step explanation:

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