an incline projection can achive maximium horizontal horizontal range at an angle of projection,ፀ=45 why?

Explanation:

For an incline projection made at angle of θ , expression of horizontal range R can be given by the following equation .

R = u² sin 2θ / g

u is initial velocity of throw .

For range R to be maximum , the value of sin 2θ must be maximum . The maximum value of sin of an angle is 1 , so

For maximum R ,

sin2θ = 1 = sin90

2θ = 90 .

θ = 45⁰ .

So for projection made at 45⁰ , horizontal range is maximum .

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