Question

if the horizontal range of projectile be a and the maximum height attained by it is B then find the velocity of projection

Answer 1

so the velocity of projection is
$\sqrt{g( {a}^{2} + 16 {b}^{2}) \div 8b }$

Answer 2

Let the velocity of the projection = u

Angle of the projection = θ

Consider a ball moving with a velocity u making an angle \theta with the horizontal plane.

On resolving the velocity, we get, $u\quad =\quad u\sin { \beta }\quad along\quad x-axis\quad and\quad u'\quad =\quad u\cos { \beta }$ along y axis.

We need find the time taken and the displacement. The displacement and time includes maximum height reached by the ball and the range of the projectile.

Time:

By using the relation: v = u + at

$\Rightarrow \quad t\quad =\quad \frac { v\quad -\quad u }{ a }$

On substituting, the condition, v = 0; u = $u\sin { \beta }$; a = -g

We get, Time taken for maximum height $=\quad t\quad =\quad \frac { u\sin { \theta } }{ g }$

Time taken for horizontal range $=\quad 2\quad \times \quad Time\quad taken\quad to\quad reach\quad maximum\quad height$

$t\quad =\quad \frac { 2\quad \times \quad u\sin { \theta } }{ g }$

Displacement:

$Range\quad =\quad a\quad =\quad \frac { { u }^{ 2 }\sin { 2\theta } }{ g } \quad =\quad \frac { { u }^{ 2 }\quad \times \quad \left( 2\sin { \theta } \cos { \theta }  \right) }{ g }\quad \longrightarrow \quad \left( 1 \right)$

$Maximum\quad height\quad =\quad B\quad =\quad \frac { { u }^{ 2 }\sin ^{ 2 }{ \theta } }{ 2g }$

${ u }^{ 2 }\sin ^{ 2 }{ \theta } \quad =\quad 2g\theta \quad \longrightarrow \quad \left( 2 \right)$

On solving equation (1)

$a\quad =\quad \frac { { u }^{ 2 }\quad \times \quad \left( 2\sin { \theta } \cos { \theta } \right) }{ g }$

$a\quad =\quad \frac { { u }^{ 2 }\quad \times \quad \left( 2\sin { \theta } \cos { \theta } \right) }{ g }$

${ a }^{ 2 }\quad =\quad \frac { { u }^{ 4 }\quad \times \quad 4\quad \times \quad \sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } }{ { g }^{ 2 } }$

$=\quad \frac { { u }^{ 2 }\cos ^{ 2 }{ \theta } }{ { g }^{ 2 } } \left( 4{ u }^{ 2 }\sin ^{ 2 }{ \theta } \right)$

$=\quad \frac { { u }^{ 2 }\cos ^{ 2 }{ \theta } }{ g } \left( 8gB \right)$

$\frac { g{ a }^{ 2 } }{ 8B } \quad =\quad { u }^{ 2 }\cos ^{ 2 }{ \theta } \quad \longrightarrow \left( 3 \right)$

On adding equation 2 and 3, we get

${ u }^{ 2 }\sin ^{ 2 }{ \theta } \quad +\quad { u }^{ 2 }\cos ^{ 2 }{ \theta } \quad =\quad 2gB\quad +\quad \frac { g{ a }^{ 2 } }{ 8B }$

$\because \quad \sin ^{ 2 }{ \theta } \quad +\quad \cos ^{ 2 }{ \theta } \quad =\quad 1$

${ u }^{ 2 }\quad =\quad 2gB\quad +\quad \frac { 2g{ a }^{ 2 } }{ 16B }$

${ u }^{ 2 }\quad =\quad 2g\quad \left( B\quad +\quad \frac { { a }^{ 2 } }{ 16B } \right)$

Thus, the velocity,

$u\quad =\quad \sqrt { 2g\quad \left( B\quad +\quad \frac { { a }^{ 2 } }{ 16B } \right) }$