Question

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​

Answer 1

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.

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