Question

A particle is projected with a velcity (u) so that its horizontal range isthrice the greatest heitht attained. What is its horizontal range?

Answer 1

Cive, ` (u^(2) sin 2 theta)/g =3 xx (u^(2) sin^(2) theta)/(2 g)`
or ` 2 sin theta cos theta =3/2 sin ^(2) theta , or tan theta =4//3,`
os ` sin theta =4//5 and cos theta =3//5`
:. Horizontal ranh `=u^(2) /g 2 theta
`= 2 u^(2))/g sin theta cos theta`
`=(2u^(2))/g xx 4/5 xx 3/5 =(24)/(25) u^(2)/g`.