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Grade: 10
        
For a projectile thrown into space with a speed v, the horizontal range is ((3)^(1/2)v^2)/2g. The vertical range is v^2/8g. The angle which the projectile makes with horizontal initially is
(a)15° (b)30° (c)45° (d)60°
11 months ago

Answers : (2)

Arun
20882 Points
							
Dear Adyasha
 
horizontal range = u^2 sin2@ / g
 
hence
 
sin2@ = sqrt(3)/ 2 
 
2@ = 60degree
 
@ = 30 degree
 
You can also check by height
11 months ago
Khimraj
2788 Points
							
max heightis given by
H = v2sin2\Theta/2g = v2/8g
so sin\Theta = ½
then \Theta = 30o
Hope it clears.............................
11 months ago
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