MHB Molly's question at Yahoo Answers regarding projectile motion

[MarkFL]

Here is the question:

A projectile is fired at an inclination of 45° to the horizontal, with a muzzle velocity of 94 feet per second?

The height h of the projectile is given by the following where x is the horizontal distance of the projectile from the firing point.

(a) How far from the firing point is the height of the projectile a maximum? Give your answer correct to the nearest foot.
(b) Find the maximum height of the projectile. Give your answer correct to the nearest foot.
(c) How far from the firing point will the projectile strike the ground? Give your answer correct to the nearest foot.

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Molly,

The equation of motion has been omitted, so I will derive it, assuming the projectile's initial height is 0 ft.

I would begin with the parametric equations of motion:

(1) $$x=v_0\cos(\theta)t$$

(2) $$y=-\frac{g}{2}t^2+v_0\sin(\theta)t$$

To eliminate the parameter $t$, we solve (1) for $t$ and substitute into (2) to get:

(3) $$y=\tan(\theta)x-\frac{g}{2v_0^2\cos^2(\theta)}x^2$$

Now we have the height $y$ of the projectile as a function of the initial velocity $v_0$, the launch angle $\theta$ and the horizontal displacement $x$.

Using the given data and known data:

$$\theta=45^{\circ}$$

$$v_0=94\,\frac{\text{ft}}{\text{s}}$$

$$g=32\,\frac{\text{ft}}{\text{s}^2}$$

We then have:

$$y=x-\frac{8}{2209}x^2$$

a) To find where the height of the projectile is a maximum, we may simply find the axis of symmetry:

$$x=-\frac{b}{2a}=-\frac{1}{2\left(-\frac{8}{2209} \right)}=\frac{2209}{16}\text{ ft}$$

b) To find the maximum height, we may now evaluate the height function at the axis of symmetry:

$$y_{\max}=y\left(\frac{2209}{16} \right)=\frac{2209}{32}\text{ ft}$$

c) To find the range $r$ of the projectile, we may double the axis of symmetry, since this axis is midway between the roots:

$$r=2\left(\frac{2209}{16} \right)=\frac{2209}{8}\text{ ft}$$