Question

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?

Answer 1

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 = (\pi)/(4)`, c)
`y_(max) = 84.375\,m`,
`t = 3.75\,s`.

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 -(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((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:

`(dx)/(d\theta) = 1350\cdot \cos 2\theta`

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

`\cos 2\theta = 0`

`2\theta = (\pi)/(2)`

`\theta = (\pi)/(4)`

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

`(d^(2)x)/(d\theta^(2)) = -2700\cdot \sin 2\theta`

`(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:

`(dy)/(dt) = 45 - 12\cdot t`

`45-12\cdot t = 0`

`t = (45)/(12)`

`t = 3.75\,s`

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

`(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`