Determine the distance from point C to the point on the horizontal surface D, if a ball is fired from a point A at the bottom of the inclined plane as shown in the figure. The mass of the ball is 4 kg ball and is of negligible size. The initial velocity of the ball is 20 m/s up the smooth inclined plane. Also, what is the velocity when it strikes the surface. 1) 36.9 m & 20 m/s 2) 32.5 m & 34 m/s 3) 42.5 m & 26 m/s 4) 61.5 m & 50 m/s Assuming the datum at A and applying the principle of conservation of energy we have : TA + VA = TB + VB 1/2(4)(20)2 + 0 = 1/2(4)(vB)2 - 4(9.81)3 vB = 18.47 m/s So, the velocity of the ball at Bis vB = 18.47 m/s The time taken for the ball to travel from B to D can be found out by the following equation: s = so + vo t + 1/2 ac t2 - 3 = 0 + 18.47 (3/5) t + 1/2 (-9.81) t2 t = 2.5 s So the time taken for the ball to reach the point D is t = 2.5 s The distance travelled by the ball horizontally to reach point D is given by the equation: s = so + vo t d = 0 + 18.47 ( 4/5) t = 36.9 m So the distance travelled by the ball horizantally to reach point D is d = 36.9 m Datum A : TA + VA = TD + VD 1/2 (4) (20)2 + 0 = 1/2 (4) (vD)2 + 0 vD = 20 m/s So the velocity with which the ball strikes the surface at D is vD = 20 m/s and also the distance travelled by it from C to reach point D, d = 36.9 m