Question

Codition for maximum harizantal range of a projectile digri

Answer 1

you must be asking that at what projectile angle the horizontal range would be maximum .

Answer:

$\alpha = 45$

Explaination:

Horizontal range=

${ u}^{2} \sin(2 \alpha ) \div g$

where

$\alpha$

represents the angle of projection with horizontal.

To make horizontal range maximum the value of

${u}^{2} \sin(2 \alpha )$

should be maximum and this is possible if the value of

$\sin(2 \alpha )$

would be maximum . The maximum value of sin is 1 when angle is of 90 degree. So

$2 \alpha = 90 \\ \alpha = 45$

Regards