The vertical height of the projectile at the time is given by `y=4t-t^(2)` and the horizontal distance covered is given by `x=3t`. What is the angle of projection with the horizontal ?

In a projectile motion, the height y = sqrt(3)t - 5t^(2) + r^(3) and horizontal distance x = t + 2t^(2) -t^(3) . The angle of projection is given by

The horizontal distance x and the vertical height y of a projectile at a time t are given by x=at and y=bt^2 +ct where a, b and c are constants The magnitude of the velocity of the projectile 1 second after it is fired 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 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 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 distances covered by a projectile at tiem t are given by x=at and y= bt^(2)+ct , wehre a,b and c are constants. What is the magnitude of the velocity of the projectile 1 second after it is fired?