A perfectly elastic ball is thrown from the foot of a plane inclined at an angle θ to the horizon . If after striking the plane at a distance ' l ' from the point of projection it rebounds and retraces its former path , find the velocity of projection in terms of l , g and θ .
-
UP 0 DOWN 0 0 1
1 Answers
1as it retraces its path clearly the ball will fall normally on the plane :P
let the angle with which it was thrown w.r.t the plane be β
we have .. vcosβ-gsinθt=0 thus t=vcosβ/gsinθ
further .. 0=vsinβt-gcosθt2/2 thus t=2vsinβ/gcosθ
further as distance =L
vcosβt-gsinθt2/2=l
solving these equations we get v = √(2lg*(cos2θ/4 + sin2θ)/sinθ
