If a projectile is fired with an initial velocity of v0...

If a projectile is fired with an initial velocity of $ v_0 $ meters per second at an angle $ \alpha $ above the horizontal and air resistance is assumed to be negligible, the its position after $ t $ seconds is given by the parametric equations $$ x = (v_0 \cos \alpha) t \quad y = (v_0 \sin \alpha)t - \dfrac{1}{2}gt^2 $$ where $ g $ is the acceleration due to gravity $ (9.8 m/s^2) $. (a) If a gun is fired with $ \alpha - 30^{\circ} $ and $ v_0 = 500\;m/s $, when will the bullet hit the ground? How far from the gun will it hit the ground? What is the maximum height reached by the bullet? (b) Use a graphing device to check your answers to part (a). Then graph the path of the projectile for several other values of the angle $ \alpha $ to see where it hits the ground. Summarize your findings. (c) Show that the path is parabolic by eliminating the parameter.

If a projectile is fired with an initial velocity of $ v_0 $ meters per second at an angle $ \alpha $ above the horizontal and air resistance is assumed to be negligible, the its position after $ t $ seconds is given by the parametric equations
$$ x = (v_0 \cos \alpha) t \quad y = (v_0 \sin \alpha)t - \dfrac{1}{2}gt^2 $$
where $ g $ is the acceleration due to gravity $ (9.8 m/s^2) $.
(a) If a gun is fired with $ \alpha - 30^{\circ} $ and $ v_0 = 500\;m/s $, when will the bullet hit the ground? How far from the gun will it hit the ground? What is the maximum height reached by the bullet?
(b) Use a graphing device to check your answers to part (a). Then graph the path of the projectile for several other values of the angle $ \alpha $ to see where it hits the ground. Summarize your findings.
(c) Show that the path is parabolic by eliminating the parameter.

Transcript

00:01 So for this first part, they tell us that we're firing some projectile at an angle of 30 degrees, and the initial velocity is 500 meters per second.

00:12 So let's just go ahead and plug these into here, and then we'll get our equation.

00:17 So first, x is going to be, so it's v .0, so it would be 500 times, so cosine of 30 degrees, so that's going to be root 3 over 2.

00:30 And then just times t.

00:33 So that would just be xizing to 250 root 3 times t.

00:40 And then for y, we'll plug this in.

00:44 Cosine of 30 is a half.

00:46 So we'd have 500 times 1 half.

00:52 T minus 1 half.

00:53 And then they tell us g is 9 .81, but i'm just going to leave it as g for right now.

00:59 So g over 2.

01:01 And then we have just t squared there.

01:08 And then that would simplify down to 250 t minus g over 2 t squared.

01:16 So these are going to be our two equations.

01:21 Now they want us to first figure out the time that it will hit the ground.

01:27 Now what this is essentially just saying is when is y equal to zero? because y is supposed to be telling us the height of it.

01:35 So we just come up here, set that equal to zero.

01:37 First thing i'm going to do is factor a t out of that.

01:40 So this is going to be t times 250 minus g over 2.

01:47 T equal to zero.

01:49 So we get t is equal to zero, or we get 250 minus g over 2.

01:56 T is equal to zero.

01:57 Well, t is equal to zero is where we're starting from, so that doesn't really make sense for us.

02:01 And then we can just solve for t.

02:04 So we would add t over and then multiply by 2 and divide by g.

02:08 So then t is going to be equal 250 times 2 is 500, and then we divide by g.

02:14 And then if we were to divide these, 500 divided by 9 .81, is approximately 50 .96 seconds.

02:26 So this is going to be our time to hit the ground.

02:33 To figure out how far away from the gun we would be, this would be us just taking what we have over here and then plugging it into x.

02:46 So this is going to be x is equal to 250 root 3.

02:52 And then i'll just use the exact answer.

02:54 So it would be 500 over g.

03:00 So 250 times 500 is 125 ,000.

03:05 So this would be 125 ,000 root 3 over g.

03:11 And then if we plug that into a calculator, that should give something around 22 ,069 .96.

03:26 And the units with this should be meters.

03:30 So that's about how far away we would be and now lastly the max height well this is just going to be what is the maximum of why so we could just take the derivative of this and then set equal to zero and since we know this is a quadratic we with a negatively in coefficient we know it's going to look something kind of like that so let's just go ahead and do that or you could just use the vertex formula as well, but i'm just going to take the derivative.

04:06 So we have dy by dt is equal to, so it would be 250 minus gt.

04:17 And then we set that equal to zero, and then we just solve for t.

04:21 So t is going to be 250 over g.

04:26 And then we can take this and plug it into y, and then that should give us our maximum value.

04:31 I mean, technically you should check by like first derivative or second derivative test to make sure that this is a maximum, but at least from like algebra, maybe pre -cal, we would know the shape of this graph already.

04:44 But yeah, so let's put that in now.

04:46 So that would be y is equal to.

04:54 So it would be 250 times 250 over g minus g over 2 and then 250 over g.

05:04 Squared.

05:08 So if we were to just multiply everything together now, so 250 times 250 divided by g which is 9 .81.

05:19 So first it's going to be approximately 6371 .5 and then minus.

05:28 So 250 divided by 9 .81 squared times 9 .81 divided by 2 .2.

05:35 So this would be minus 3 ,000.

05:37 85 .5 .2 and then if we subtract these, we should get something around 3 ,185 .53.

06:00 And then again, this should be in meters.

06:03 So this is going to be the max height that we reach.

06:09 Now for part b, they want us to sketch the graph at alpha is equal to 30 degrees.

06:18 And then just check to see if these match up.

06:20 So what we're really looking for is that the maximum distance away is this 22 ,000, and then the maximum height is 31, or not 31, 3 ,100, or about 3 ,200.

06:32 So i went ahead and already did this in desmos.

06:37 So you can see i kind of have this rope exactly as the way they had it before.

06:41 So if you want to ever do a parametric equation, all you really need to do is put like a coordinate or some parentheses.

06:49 And then like, let's say i want to do cosine t, sign t...

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