Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 2 (Q.No. 42)
42.
If the horizontal range is 2.5 times the greatest height, the angle of projection of the projectile, is
Discussion:
4 comments Page 1 of 1.
Nizam alizai said:
4 years ago
For three times the height, the angle of projection should be 53 degrees.
Aswathy said:
6 years ago
Horizontal range,R= (u^2sin(2$))/g........(eq1)
Maximum height,H=(u^2sin^2($))/2g........(eq2)
Here,
Horizontal range= 2.5*Maximum height
R=2.5*H.....(eq3)
Substitute eq 1 & eq 2 in eq3
Sin(2$)=(2.5sin^2($))/2
2sin($)cos($)= (2.5sin^2($))/2
tan($)=1.6
$=57.99°=58°
Maximum height,H=(u^2sin^2($))/2g........(eq2)
Here,
Horizontal range= 2.5*Maximum height
R=2.5*H.....(eq3)
Substitute eq 1 & eq 2 in eq3
Sin(2$)=(2.5sin^2($))/2
2sin($)cos($)= (2.5sin^2($))/2
tan($)=1.6
$=57.99°=58°
(1)
Hardik said:
6 years ago
@Oyewale abe. From that equation.
How can we find are and θ at the same time?
How can we find are and θ at the same time?
Oyewole ABE said:
9 years ago
The relationship is,
H = (Rtanθ)/4 hence 58° is correct.
H = (Rtanθ)/4 hence 58° is correct.
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