A ball is projected from a cliff of height `h = 19.2` m at an angle a to the horizontal. It hits an incline passing through the foot of the cliff, inclined at an angle `theta` to the horizontal. Time of flight of the ball is `T = 2.4 s`. Foot of the cliff is the origin of the co-ordinate system, horizontal is x direction and vertical is y direction (see figure). Plot of y co-ordinate vs time and y component of velocity of the ball `(v_(y))` vs its x co-ordinate (x) is as shown. x and y are in m and time is in s in the graph. `[g = 10 m//s^(2)]`


(a) Find the angle of projection a
(b) Find the inclination `(theta)` of the incline.
(c) If the ball is projected with same speed but at an angle `theta (=` inclination of incline) to the horizontal, will it hit the incline above or below the point where it struck the incline earlier?

Answer

(a) `alpha = tan^(-) ((3)/(4))` (b) `theta = tan^(-) (1)/(2)` (c) The ball will hit at a point lower than the earlier spot.