MHB Molly's question at Yahoo Answers regarding projectile motion

AI Thread Summary
The projectile is launched at a 45° angle with a muzzle velocity of 94 feet per second. The maximum height occurs at a horizontal distance of approximately 138.06 feet from the firing point, yielding a maximum height of about 69.03 feet. The projectile will strike the ground at a distance of approximately 276.12 feet from the firing point. The calculations utilize the parametric equations of motion and the derived height function based on the initial conditions. The discussion emphasizes the importance of understanding projectile motion in physics.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
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.
 
Mathematics news on Phys.org
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:

$$r=2\left(\frac{2209}{16} \right)=\frac{2209}{8}\text{ ft}$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top