Physics, asked by aswins267, 8 months ago

The ratio of kinetic energy of a projectile at the point

of projection to the maximum height is 4 : 1. The

ratio of maximum height to the range is​

Answers

Answered by Anonymous
11

Answer:

 \boxed{\sf \frac{Maximum \ height \ (H_{m}) }{Range\ (R)}  = \frac{\sqrt{3}}{4} }

Given:

Ratio of kinetic energy of a projectile at the point of projection to maximum height = 4:1

To Find:

Ratio of maximum height to the range

Explanation:

Let initial velocity at the point of projection be 'v'

Velocity at maximum height is equal to horizontal velocity i.e. v cosθ

(θ = Angle of projection)

According to the question;

\sf  \implies \frac{K.E._{Point \ of \ projection} }{K.E._{Maximum \ height} } = \frac{4}{1}  \\  \\ \sf  \implies  \frac{ \cancel{ \frac{1}{2} m }{v}^{2} }{ \cancel{ \frac{1}{2}m} {(v \: cos \theta)}^{2}  }  =  \frac{4}{1}  \\  \\ \sf  \implies  \frac{ \cancel{ {v}^{2} }}{  \cancel{{v}^{2}  }{cos}^{2}  \theta}  =  \frac{4}{1} \\  \\  \sf  \implies  {cos}^{2}  \theta =  \frac{1}{4}  \\  \\ \sf  \implies cos \theta =  \sqrt{ \frac{1}{4} }  \\  \\ \sf  \implies cos \theta =  \pm \frac{1}{2}  \\  \\  \\  \therefore   \\ \sf \implies \theta = 60^{ \circ}  \: or \:120 ^{\circ }

\sf Maximum \ Height \ (H_{m}) =  \frac{ {v}^{2}  {sin}^{2} \theta }{2g}  \\  \sf Range \ (R) =  \frac{ {v}^{2}sin 2 \theta}{g}

 \sf  \frac{H_{m}}{R} =  \frac{ \frac{  \cancel{{v}^{2} }{sin}^{2} \theta  }{2 \cancel{g}} }{ \frac{ \cancel{ {v}^{2}}sin2 \theta }{ \cancel{g}} } \\  \\   \sf = \frac{ \frac {{sin}^{2} \theta  }{2} }{ sin2 \theta  } \\  \\  \sf  = \frac{ {sin}^{2} \theta }{2(sin2 \theta)} \\  \\   \sf =  \frac{ \cancel{sin \theta} \times sin \theta}{2 \times 2 \cancel{sin \theta}cos \theta}   \\  \\ \sf =  \frac{sin \theta}{4cos \theta}  \\  \\  \sf =  \frac{tan \theta}{4} \\  \\   \sf =  \frac{tan60^{ \circ }}{4}  \:  \:  \: or \:  \:  \:  \frac{tan120^{ \circ}}{4}  \\  \\  \sf =  \frac{ \sqrt{3} }{4}  \:  \:  \: or \:  \:  \:  -  \frac{ \sqrt{3} }{4}

As it's ratio of distance so it can't be negative.

 \therefore \\  \frac{H_{m}}{R} =  \frac{ \sqrt{3} }{4}

Attachments:

Anonymous: Awesome ^^"
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