What is the angle of projection at which horizontal range and maximum height are equal
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Answered by
1
at tan theta=4..the horizontal height and range of the projectile are equal
Arya18Pandey:
u can equate the formula for maximum height and the range of the projectile to the the final answer as tan theta =4
Answered by
6
Horizontal range = u^2 sin2theta/g
Maximum height = u^2 sin^2theta/2g
So according to the question,
u^2 sin^2theta/2g = u^2 sin2theta/g
=> sin^2theta/2 = 2 sin theta cos theta
=> sin theta = 4 cos theta
=> sin theta/cos theta = 4
=> tan theta = 4
=> theta = tan^-1(4)
Maximum height = u^2 sin^2theta/2g
So according to the question,
u^2 sin^2theta/2g = u^2 sin2theta/g
=> sin^2theta/2 = 2 sin theta cos theta
=> sin theta = 4 cos theta
=> sin theta/cos theta = 4
=> tan theta = 4
=> theta = tan^-1(4)
hope you understand
$
{\tan}\hspace{0.33em}\mathit{\theta}
$
= 4
{\tan}\hspace{0.33em}\mathit{\theta}
$
= 4
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