Correct Answer - D
`R = (u^(2)sin 2 theta)/(g) rArr sqrt(3) = ((sqrt(20))^(2)sin 2 theta)/(10)`
`sin 2 theta = (sqrt(3))/(2) rArr 2 theta = 60^(@), 120^(@)`
`rArr theta = 30^(@), 60^(@)`
(A) For `theta = 30^(@)`
`H = (v^(2)sin^(2) theta)/(2g) rArr H = ((sqrt(20))^(2)(1//2)^(2))/(2xx10) = (1)/(4) = 0.75m`
(B) At heghiest point of motion `u` is minimum
`u_(min) = u cos theta = sqrt(20) xx cos 30^(@) = sqrt(15) m//s`
(C ) `T = (2u sin theta)/(g)`
For `theta = 30^(@), T = (2sqrt(20)xx(1)/(2))/(10) = (1)/(sqrt(15))`
For `theta = 60^(@) , T = (2xxsqrt(20))/(10) xx (sqrt(3))/(2) = sqrt((3)/(5))sec`.
(D) `U_(max) = mgH = 1 xx 10 xx 0.25 = 2.5 J`