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For a projectile thrown up an incline plane (making an angle theta with horizontal ),maximum range on the incline can be achieved by projecting at an angle .................... From the horizontal (theta is acute).

Suraj Kumar

6 Years agoGrade

2 Answers

Arun

6 Years ago

Dear student

As we know

range= u² sin 2 theta/g

Hence for maximum range sin2 theta = 1

Hence 2 theta = 90 degree

Hence theta = 45°

Vikas TU

6 Years ago

Dear student

R = u^2Sin2Q /g

Rmax = u^2 /g

So, Sin2Q =1

2Q = Sin^-1 (1)

2Q = 90 degree

Q = 45 degree

Hope this helps

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