45* maximum range projectile? WHY?

[00945584]

Why is the range for a projectile max at 45 degrees?
The equation for range is vcos*t= range.
As cosine increases
0= 1
30= .9
45= .7
60= .5
90= 0

so as the angle increases the range decreases? but that is wrong as I know but why mathmatically. Princeton review says the range is this formula.

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[sv3]

Why is the range for a projectile max at 45 degrees?
The equation for range is vcos*t= range.
As cosine increases
0= 1
30= .9
45= .7
60= .5
90= 0

so as the angle increases the range decreases? but that is wrong as I know but why mathmatically. Princeton review says the range is this formula.

i'll take a stab at this and i think its a compromise between Vy and Vx. If its a 30degree angle, sure the cosq is greater but the vertical velocity will be low so that time in air is low (range is dependent on time). And if vertical velocity is greater than 45degrees, then your horizontal velocity will be lower. So i think 45degrees allows you to have the best combination of time in air plus horizontal velocity. now only if i could prove that mathematically for you, but alas, i cant!

steve

 

[00945584]

is my range formula still applicable to all mcat questions then?

 

[TraderDoc]

The ideal theta is 45 degrees because this is the perfect compromise between flight time (y-comp) and horizontal speed (x-comp). Sin45 = Cos45 . At Cos 0=1, you have maximum horizontal velocity but no flight time. Therefore you will get a bowling effect rather than a tossing effect which has greater displacement.

 

[00945584]

so i have to be selective with vcos*= range?

if this is the case it screws up how i calculate range at angles? this gets me even more confused.

i understand that 45* is the max range, but now calculating range is a problem of a projectile.

im getting lost.

 

[sv3]

so i have to be selective with vcos*= range?

if this is the case it screws up how i calculate range at angles? this gets me even more confused.

i understand that 45* is the max range, but now calculating range is a problem of a projectile.

im getting lost.

dont sweat it. your range formula always applies. i havent come across an instant where it doesn't. range always = vx (vcosq) x time

 

[starbaduk]

Why is the range for a projectile max at 45 degrees?
The equation for range is vcos*t= range.
As cosine increases
0= 1
30= .9
45= .7
60= .5
90= 0

so as the angle increases the range decreases? but that is wrong as I know but why mathmatically. Princeton review says the range is this formula.

That is the correct equation for range. You have to remember that t is also a function of theta. So as you change theta from 0 to 45, it is true the value of v cos theata would decrease (as you noted), but t would increase.

 

[starbaduk]

I knew 45 degrees gave you max range, but I was curious. So I worked it out:

y = v sin theta * t + 1/2 at^2
0 = v sin theta * t + 1/2 at^2
0 = t(v sin theta + 1/2 at)

Non trivial solution:
v sin theta + 1/2 at = 0
t = - 2 v sin theta / a

So range:
x = v cos theta * t
x = v cos theta * - 2v sin theata / a
= - 2v^2/a * cos theta * sin theta

Basically you want to maximize cos theta * sin theta:

f = cos theta * sin theta

then max value of y can be found by taking the derivative

f' = cos^2 theta - sin^2 theta

0 = cos ^2 theta - sin^2 theta
sin^2 theta = cos^2theta

45 degrees one of the solutions that satisfy above equation

 

[MBHockey]

Just think about it conceptually. It always helps to think of extreme cases.

Why would it have to be 45 degrees, an exact compromise between full vertical and full horizontal?

The vertical component of velocity determines the time it is in the air. The horizontal component determines how far it will go. And you want to maximize both of these values simultaneously.

Why? Because if your try to go for a really long time of flight by just shooting vertically, you won't get anywhere horizontally. Similarly, if you want to go for the longest range and just shoot it completely horizontally, your time in the air is zero, so you've only screwed yourself. The happy compromise is at 45 degrees.