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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).

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).

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2 Answers

Arun
25750 Points
4 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
14149 Points
4 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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