So far I have used the Product-to- sum Formula :sincos The range of a projectile

Question

So far I have used the Product-to- sum Formula :sincos The range of a projectile propelled downward from the top of an inclined plane at an angle theta to theinclined plane is given by when the projectile hits the inclinedplane. Here vo is the initail velocity ofthe projectile , phi is the angle the plane makes withrespect to the horizontal, and g is the accleration due togravity. (a) Show that for fixed voand phi the maximum range down the incline is givenby So far I have used the Product-to- sum Formula :sin alpha cos beta and have ended up with the next thing is to say that sin(2 theta - phi) =sin(pi/2)=1 this is how I think this willwork to cancel out (1+sin phi) but I don't know how this isdone? Thank you. Comment

Explanation / Answer

For R() to be maximum its first deriavative with respectto should be zero Since, vo , g, are constant d R() = 2vo2.        d { sin(2 -) + sin} = 0    d       gcos2               d which becomes cos (2 - ) = 0 this implies that (2 - ) = /2 and therefore, sin (2 - ) = sin /2 = 1 Now you can cancel and get the required expression.

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