Optimal angle for a projectile part 4 Finding the optimal angle and distance with a bit of calculus

[SuperHempo]

Question: Why dont you use the fact that the horizontal speed is constant? The distance is Sh*t. So you have to calculate when the time that the object is in the air, given a speed m, is maximal.

Then you have: S * sin (teta)*t = 1/2 g t^2

This gives t = 2*S*sin(teta)/g.

This is maximal when sin(teta) = 1, because S en g are constants. Sin(teta)= 1 when teta is 1/4 pi.

This is way faster :-)

[SuperSonic44]

Good video but as has been pointed out you need to use the chain rule because that only works for radians. Also it would have been easier to use the identity sin(2x) = 2sin(x)cos(x)

[olkomat1]

@SuperSonic44 actually i wouldn't have used derivatives at all. Just think about maximizing sin(2x); the sine function goes from -1 to 1 so maximizing it means when does it equal 1? sin(90degrees) is 1, and compare this to sin(2x) you see that 2x=90degrees, so x=45degrees :)

[MarvelsofaLifetime]

All that work to distill and get something as simple as (s^2/g) – simply fascinating. This is why I love physics so much.

[manshet1]

god bless you man!

[Stratocaster1111]

i like the new HD videos. Is it a new tablet?

[RiceHawt]

lol he sounds so mad in this video

[sa78223]

i love you sal <3

[sa78223]

seriously it is so dumb to criticize a person, that provides us with such great videos and that most likely understands the subject better than you do, about such a trivial thing...this is almost as ridiculous as people talking shit cus he uses 10 for gravity instead of 9.8

the end couldve been manipulated in numerous ways, there really is no wrong or right way...choose whats most comfortable for you and leave it at that.

[fiercex]

Don't know why everyone is fussing over the algebraic manipulation at the end of this problem. There are many ways to solve this, and as always, Sal chooses to show at least one of the more intuitive ways.

[hppy2cu]

Tedious.

If cos^2 = sin^2, cos = +/- sin, by factoring the quadratic [difference of squares]. both positive, refer to unit circle in first quadrant.

Oh yes, d/d(theta) sin(theta) = Pi/180 cos(theta), if theta is measured in degrees.

This is a very disappointing video.

[dalcde]

I though d/dx(sinθ) = cosθ is only true for θ in radians, or it should be π/180(cosθ), but since you could divide by π/180, the answer is still correct.

[megaelliott]

OMG, I found this same thing out not too long ago, except my proof of the optimal angle didn't involve tangents. At the point where you had cos(theta)*sin(theta), I then used the law that cos(theta)*sin(theta) = sin(2*theta)/2, and thus sin(2*theta) is greatest when 2*theta is 90 degrees, or theta is 45 degrees.

[madah012]

This is excellent work but I can suggest that you put this alternative method starting from d = s^2/g 2sin(theta)cos(theta) since sin(2theta)=2sin(theta)cos(the­ta) d= s^2/g sin(2theta) d'(theta)= s^2/g cos(2theta) * 2 2s^2/g cos(2theta)=0 cos(2theta)=0 cos(90)=0 cos(2theta)=0 2theta=90 theta = 45 d=s^2/g sin(2theta) d=s^2/g *1=s^2 / g

[norwayte]

Physics can be fun.

[bazza4288]

awsome video

how would you go about finding the optimal angle for a set distance?

[frecuenciajuaniyo]

excellent videos... very good demonstration..

[abewell19]

What if the starting height is not 0? The max angle calculations get much hairier.