# Thread: max range of projectile

1. ## max range of projectile

a particle is projected up an inclined plane with initial velocity u m/s .the line of projection makes angle a with the horizontal and the inclined plane makes angle b with the horizontal. show that the range is a maximum when a=45°+b/2

so far I got the time of flight =$\displaystyle 2usinb-a/(gcosa)$
and i have gotten the range =$\displaystyle [cos(b-a) -{(sina*sinb-a)/cosa} ] [(2u^2sinb-a)/gcosa]$
from here i cant see how to maximise this using differentiation or trigonometry.

2. ## Re: max range of projectile

if you can not understand my brackets, sorry and please ask me to explain what you do not get

3. ## Re: max range of projectile

Originally Posted by markosheehan
a particle is projected up an inclined plane with initial velocity u m/s .the line of projection makes angle a with the horizontal and the inclined plane makes angle b with the horizontal. show that the range is a maximum when a=45°+b/2

so far I got the time of flight =$\displaystyle 2usinb-a/(gcosa)$
and i have gotten the range =$\displaystyle [cos(b-a) -{(sina*sinb-a)/cosa} ] [(2u^2sinb-a)/gcosa]$
from here i cant see how to maximise this using differentiation or trigonometry.
Since the problem is to show that "a= 45+ b/2", a is the variable and you can set the derivative, with respect to a, equal to 0.

4. ## Re: max range of projectile

there is more than one variable (a and b) so i do not see how i can do this.
I am trying to get a as one term and then differentiate it.
i have gotten here : [cosa * cosb-a -(sinasinb-a)] [(2u^2 sinb-a)/g]
and then using the trig formula cos(A+B) = cosAcosB− sinAsinB
cos a+b-a [(2u^2 sinb-a)/g]
cos b [(2u^2 sinb-a)/g]
I can not get the answer

5. ## Re: max range of projectile

show that the range is a maximum when a=45°+b/2
note incline angle $b$ may be treated as a constant ... angle $a$ is variable.

Let the surface of the incline be the x-axis, y-axis perpendicular to the x-axis, $(x_0,y_0) = (0,0)$ ... see attached diagram

$x = u\cos(a-b) \cdot t - \dfrac{1}{2}g\sin{b} \cdot t^2$

$y = u\sin(a-b) \cdot t - \dfrac{1}{2}g\cos{b} \cdot t^2$

$y = 0 \implies t = \dfrac{2u\sin(a-b)}{g\cos{b}}$

substituting for $t$ in the range equation, $x$, yields ...

$x = \dfrac{u^2}{g\cos^2{b}}\bigg[\cos{b} \cdot 2\cos(a-b)\sin(a-b)-\sin{b} \cdot 2\sin^2(a-b)\bigg]$

using double angle identities ...

$x = \dfrac{u^2}{g\cos{b}}\bigg[\cos{b}\sin[2(a-b)]-\sin{b}+\sin{b}\cos[2(a-b)]\bigg]$

$\dfrac{dx}{da} = \dfrac{2u^2}{g\cos{b}}\bigg[\cos{b}\cos[2(a-b)]-\sin{b}\sin[2(a-b)]\bigg]$

using the sum identity for cosine ...

$\dfrac{dx}{da} = \dfrac{2u^2}{g\cos{b}} \cdot \cos[b +2(a-b)] = \dfrac{2u^2}{g\cos{b}} \cdot \cos(2a-b)$

$\dfrac{dx}{da}=0 \implies \cos(2a-b)=0 \implies 2a-b = 90^\circ \implies a = 45^\circ + \dfrac{b}{2}$

6. ## Re: max range of projectile

I am having trouble understanding some of the double angle identities.
i can not see how 2sin^2 (a-b) goes to sinb*cos[2(a-b)]

i also can not see how u^2/gcosb [cosb*sin2(a-b) -sinb*+sinbcos2(a-b) goes to
2u^2/gcosb [cosb*cos2(a-b) -sinb*sin2(a-b)
i can not see how you are factorizing by 2

thanks

7. ## Re: max range of projectile

$\cos (2\theta) = 1-2\sin^2 \theta$
$2\sin^2(a-b) = 1-\cos (2(a-b))$

So, $x = \dfrac{u^2}{g\cos b}\left(\cos b \sin(2(a-b)) - \sin b + \sin b \cos(2(a-b))\right)$

Then, skeeter took the derivative:

$\dfrac{dx}{da} = \dfrac{u^2}{g\cos b}\left[\cos b (2)\cos(2(a-b)) - 0 + \sin b (-2)\sin(2(a-b))\right]$

skeeter simplified:

$\dfrac{dx}{da} = \dfrac{2u^2}{g\cos b}\left[\cos b \cos(2(a-b)) - \sin b \sin(2(a-b))\right]$