We will show later in the text that if a projectile is fired...

We will show later in the text that if a projectile is fired from ground level with an initial speed of $v_{0}$ meters per second at an angle $\alpha$ with the horizontal, and if air resistance is neglected, then its position after $t$ seconds, relative to the coordinate system in the accompanying figure is $$ x=\left(v_{0} \cos \alpha\right) t, \quad y=\left(v_{0} \sin \alpha\right) t-\frac{1}{2} g t^{2} $$ where $g \approx 9.8 \mathrm{~m} / \mathrm{s}^{2}$ (a) By eliminating the parameter, show that the trajectory lies on the graph of a quadratic polynomial. (b) Use a graphing utility to sketch the trajectory if $\alpha=$ $30^{\circ}$ and $v_{0}=1000 \mathrm{~m} / \mathrm{s}$ (c) Using the trajectory in part (b), how high does the shell rise? (d) Using the trajectory in part (b), how far does the shell travel horizontally?

Transcript

00:01 So we're given two equations here, both functions of time, and they describe a projectile's motion through space.

00:11 And what we're asked to do is to eliminate the dependence on time and graph it and then analytically solve for the height and the range.

00:22 Well, i've already graphed it, as you can see here on the top right.

00:24 And i have the range over here, obviously, it's 88 ,000 roughly, and the height is roughly 13 ,000.

00:36 So to get those values, the first thing i'm going to do is eliminate the parameter of time.

00:42 If you do that, i'm going to start with four is i'm going to put in here.

00:46 X equals b0, cosine of beta, t.

00:52 And i'm just going to solve this for t, which is t equals t.

00:57 Equals x over b0 cosine of theta.

01:03 Now, all we have to do is take this value that we just solve for t and plug it into our vertical equation y.

01:13 And that gives us y as a function of x equals y0 plus v0 times our replacement for t is x over b0 times cosine xx5x2 minus g over 2 times x over v0 cosine theta and that is squared so the first thing we want to do is find the height and the equation is at its highest point when it's vertical velocity v y is equal to zero so we can take that equation v y which equals v y gt which equals v0 and the sign of theta minus g times x over v0 times the cosine of theta and again that equals zero and we want to solve this for x so we can begin by adding g x over b0 cosine to both sides gives us b z 0 times the sine theta equals g times x over v0 cosine data.

02:57 And now we can solve for x by multiplying both sides by v0 cosine data and dividing both sides by g.

03:05 And we get x equals b0 squared times the sine data times the cosine of theta divided by by g.

03:24 And i'll rewrite that with the values inserted.

03:31 So that equals 1 ,000 meters per second.

03:36 That value is squared times the sign of 30 degrees, which is a square root of one quarter, or just one half, times the cosine of 30 degrees, which is the square root of three quarters.

03:52 And that is divided by 9 .8...

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