The equation of trajectory of an oblique projectile is `y=x-(1)/(2)x^(2)` .The time of flight of projectile will be.
A) `2/ sqrt g`
B) `3/ sqrt g`
C)` 4/ sqrt g`
d) `2 sqrt g/g`

The equation of trajectory of an oblique projectile y = sqrt(3) x - (g x^(2))/(2) . The angle of projection is

The equation of trajectory ofa projectile is y = 10x-5/9x^2 If we assume g = 10 ms^(-2) , the range of projectile (in metre) is

The trajectory of a projectile in a vertical plane is y=sqrt(3)x-2x^(2) . [g=10 m//s^(2)] Time of flight of the projectile is :-

The trajectory of a projectile in a vertical plane is y=sqrt(3)x-2x^(2) . [g=10 m//s^(2)] Angle of projection theta is :-

The trajectory of a projectile in a vertical plane is y=sqrt(3)x-2x^(2) . [g=10 m//s^(2)] Maximum height H is :-

The trajectory of a projectile in a vertical plane is y=sqrt(3)x-2x^(2) . [g=10 m//s^(2)] Range OA is :-

Trajectory of particle in a projectile motion is given as y = x - (x^(2))/(80) . Here x and y are in metre. For this projectile motion match the following with g = 10 ms^(-2) .

The shortest distance between line y-x=1 and curve x=y^2 is (a) (3sqrt2)/8 (b) 8/(3sqrt2) (c) 4/sqrt3 (d) sqrt3/4

Three numbers form an increasing G.P. If the middle number is doubled, then the new numbers are in A.P. The common ratio of the G.P. is (A) 2-sqrt(3) (B) 2+sqrt(3) (C) sqrt(3)-2 (D) 3+sqrt(2)

IF f(x) = sqrt(x) and the composite function f(g(x))= sqrt(4x^2-5) , which of the following could be g(x) ?