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?

A ball is projected from a cliff of height h = 19.2 m at an angle a to

1.	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» Answer :(a) `ALPHA = TAN^(-) ((3)/(4))` (b) `theta = tan^(-) (1)/(2)` (c) The ball will HIT at a point lower than the earlier SPOT.