Question

8
For a projectile fired with a certain velocity, the maximum possible range and the maximum
height attainable are related as​

Answer 1

Projectile is thrown for maximum range.

• It means that the angle of projection is 45° (i.e. maximum range is attained at this angle.)

Now, range will be :

$R = \dfrac{ {u}^{2} \sin (2 \theta) }{g}$

$\implies R = \dfrac{ {u}^{2} \sin ( {90}^{ \circ} ) }{g}$

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

Now, max height will be :

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

$\implies H= \dfrac{ {u}^{2} {\sin}^{2} ( {45}^{ \circ} ) }{2g}$

$\implies H= \dfrac{ {u}^{2} {( \frac{1}{ \sqrt{2} } )}^{2} }{2g}$

$\implies H= \dfrac{ {u}^{2} }{4g}$

$\implies H= \dfrac{1}{4} \times \dfrac{ {u}^{2} }{g}$

$\implies H= \dfrac{1}{4} \times R$

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

Hope It Helps.