Question

Find an expression for maximum range of a particle which is projected with an initial velocity of ‘u’ inclined at an angle of ‘β’ with the horizontal.

Answer 1

Consider a particle performing projectile motion.
R – Horizontal Range
T – Total flight time
Considering vertical components of motion,

`s =ut +1/2 at^2`

`0 " usin "(beta)-1/2 gT^2`

`T = (2usin(beta))/g`

Considering horizontal components of motion,

`s =ut +1/2at^2`

R = ucos(β)T + 0 …….(as acceleration in x direction is zero)

`"R = ucos"(beta)xx(2usin(beta))/g`

`R=(u^2sin(2beta))/g`

For maximum Range, sin(2β) should be maximum, i.e. sin(2β) = 1, i.e. 2β = 90, i.e. β=45⁰

`R_(max)= u^2/g`