Question

If both horizontal range and Hmax are equal than find horizontal range?

Answer 1

Given:

Horizontal range (R) = Maximum height $\sf (H_{max})$

To Find:

Horizontal range (R)

Answer:

Horizontal range:

$\boxed{ \bf{R = \dfrac{ 2{u}^{2}sin \theta cos \theta }{g} }}$

Maximum height:

$\boxed{ \bf{H_{max} = \dfrac{ {u}^{2} {sin}^{2} \theta }{ 2g} }}$

So,

$\rm \implies R = H_{max} \\ \\ \rm \implies \dfrac{ 2 \cancel{{u}^{2}} \cancel{sin \theta }cos \theta }{ \cancel{g}} = \dfrac{ \cancel{{u}^{2} }{sin}^{ \cancel{2}} \theta }{ 2 \cancel{g}} \\ \\ \rm \implies 2cos \theta = \dfrac{sin \theta}{2} \\ \\ \rm \implies \dfrac{sin \theta}{cos \theta} = 2 \times 2 \\ \\ \rm \implies tan \theta = 4$

Thus,

$\rm sin \theta = \dfrac{4}{ \sqrt{17} } \\ \rm cos \theta = \dfrac{1}{ \sqrt{17} }$

By substituting values of sinθ & cosθ in the equation of horizontal range we get:

$\rm \implies R = \dfrac{ 2{u}^{2} }{g} \times \dfrac{4}{ \sqrt{17} } \times \dfrac{1}{ \sqrt{17} } \\ \\ \rm \implies R = \dfrac{2 {u}^{2} }{10} \times \dfrac{4}{17} \\ \\ \rm \implies R = \dfrac{8{u}^{2} }{170} \\ \\ \rm \implies R = \dfrac{4{u}^{2} }{85} \: m$

$\therefore$ $\boxed{\mathfrak{Horizontal \ range \ ( R ) = \dfrac{4{u}^{2} }{85} \: m}}$