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 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 y and 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)) meter and x = 6t meter , where t is in second. The velocity with which the projectile is projected is

The height y and the distance x along the horizontal plane of a projectile on a certain planet [with no surrounding atmosphere] are given by y=[5t-8t^2] metre and x = 12t metre where t is the time in second. 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 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 horizontal and vertical displacements x and y of a projectile at a given time t are given by x= 6 "t" and y= 8t -5t^2 meter.The range of the projectile in metre is:

The equations of motion of a projectile are given by x=36tm and 2y =96t-9.8t^(2)m . The angle of projection is

The horizontal and vertical displacements of a projectile are given as x=at & y=bt-ct^(2) .Then velocity of projection is