The horizontal distance x and the vertical height y of a projectile at time t are given by x = a tandy = bt^(2)+ct wherea, b and c are constants. Then

The horizontal distance x and the vertical height y of a projectile at

1.	The horizontal distance x and the vertical height y of a projectile at time t are given by x = a tandy = bt^(2)+ct wherea, b and c are constants. Then
Answer» The speed of the projectile 1 second after it is fired is `(a^(2)+b^(2)+C^(2))^(1//2)` The angle with the horizontal at which the projectile is fired is `tan^(-1)(c//a)` The ACCELERATION due to gravity is `-2B` The INITIAL speed of the projectile is `(a^(2)+c^(2))^(1//2)` ANSWER :B::C::D