A particle is projected with velocity root 2gh such that it crosses two walls of height h, separated by h. what is the angle of projection?
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Velocity √2gh
So, R= v^2.sin2©/g = √2gh
Height of walls= H=v^2.sin^2©/2g=h
Dividing equation R/H
V^2.sin2©/g/v^2.sin^2©/2g
2sin2©/sin©=√2g
4cos©/sin©=√2g
tan©=4√2g
=2×√2/√2g
= 2/g
Its also value of R,max
We know maximum value R is 45°
So, tan©= 45°. Answer..
I hope you will agree with this.....
So, R= v^2.sin2©/g = √2gh
Height of walls= H=v^2.sin^2©/2g=h
Dividing equation R/H
V^2.sin2©/g/v^2.sin^2©/2g
2sin2©/sin©=√2g
4cos©/sin©=√2g
tan©=4√2g
=2×√2/√2g
= 2/g
Its also value of R,max
We know maximum value R is 45°
So, tan©= 45°. Answer..
I hope you will agree with this.....
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