The vertical height y and horizontal distance x of a projectile on a certain planet are given by `x= (3t) m, y= (4t-6t^(2))` m where t is in seconds, Find the speed of projection (in m/s).

The height y and distance x along the horizontal plane of a projectile on a certain planet are given by x = 6t m and y = (8t^(2) - 5t^(2))m . The velocity with which the projectile is projected is

The height y and distance x along the horizontal plane of a projectile on a certain planet are given by x = 6t m and y = (8t - 5t^(2))m . The velocity with which the projectile is projected is

The height y nad the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by y = (8t - 5t^2) m and x = 6t m , where t is in seconds. The velocity with which the projectile is projected at t = 0 is.

The height y nad the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by y = (8t - 5t^2) m and x = 6t m , where t is in seconds. The velocity with which the projectile is projected at t = 0 is.

The height and horizontal distance of a projectile are y= 8t-5t^2 m and x=6t m. its projection velocity is

The height y and the distance x along the horizontal plane of projectile on a certain planet (with no atmosphere on it) are given by y = (8t-5t^(2)) metre and x = 6t metre , where time t is in seconds. Find the initial velocity with which the projectile is projected.

The height and horizontal distances covered by a projectile in a time t are given by y=(4t-5t^(2)) m and x=3t m the velocity of projection is (A) 4ms^(-1) (B) 3ms^(-1) (C) 5ms^(-1) (D) 10ms^(-1)