Question

if R is maximum horizontal range of a particle then wat would be the greatest height attained by it?

Answer 1

range is maximum when particle is projected at angle 45° with horizontal.

so, R =
$\frac{ {u}^{2} }{g}$
Hmax =
$\frac{ {u}^{2} { \sin ^{2} ( \ \alpha ) } }{2g} =$
= 0.5 x R x (1/2)^2

= R/4. ans

Answer 2

Explanation:

Let R is the maximum horizontal range of the projectile. The formula of the range is given by :

$R=\dfrac{u^2\ sin2\theta}{g}$

For maximum range,

$sin2\theta=1\\\theta=45^{\circ}$

Putting the value of $\theta$ in above equation, we get :

$R=\dfrac{u^2}{g}$

The formula for the maximum height attained is given by :

$H=\dfrac{u^2\ sin^2\theta}{2g}$

$H=\dfrac{1}{2}\times \dfrac{u^2}{g}\times sin^2(45)$

$H=\dfrac{1}{2}\times R\times sin^2(45)$

$H=\dfrac{R}{4}$

So, the greatest height attained by the particle is R/4. Hence, this is the required solution.