Here, `theta =30 ^(@), u=20 m//s`
(a) Time taken by ball to reach the highest point, `t T/2 =(u sin theta)/g =(20)/(10) xx isn 30^(0) =xx (1//2) =1 s`
(b) The maximum height `=(u^(2) sin ^(2) xx theta)/(2g) =((20)^(2) xx sing ^(2) xx sin ^(2) 30^(0))/(2xx10) =5m`
(c ) The borixontal rane `=(u^(2) sin ^(2) xx theta)/(g) =((20)^(2) xx sing ^(2) xx sin ^(2) 30^(0))/(10) =34.64 m.
(d) The time of flotht `=(2 usin) theta)/g =(2 xx 20xx sin 30^(0))/(10) =2 s.