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Grade: 12th pass
        
Please provide explanatory answer and reply soon...
one month ago

Answers : (1)

Arun
22021 Points
							
This is a question on horizontal projection. 
Let R be the horizontal distance covered by the projectile then and a be the angle of projection. 
U be the initial velocity and in our case it is 16m/s
And g be the acceleration due to gravity and in our case it is 10.
​​
Then R = U² Sin 2a / g
R = 14.4 + b (it hits the roof half way and the width of the roof is 2b)
R = 16² sin 2(45°) / 10
= (256 × 0.98769) / 10 = 25.28m
b = 25.28 - 14.4 = 10.88m
2b = 2 × 10.88 = 21.76m
This is the width of the building. 
The angle of projection when initial velocity is 10.3m/s and the projectile crosses the roof is given by :
Height of the roof equals U Sin a where u is the initial velocity.
4.8 = 10.3 × Sin a
Sin a = 4.8 ÷ 10.3 = 0.4660
Sin ⁻¹0. 4660 = 30.86°
 
 
one month ago
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