Let the basket ball be launched from origin with initial velocity of #vecu# which makes an angle #theta# with #x#-axis.
Also assuming negligible friction, the distance traveled in #x# direction is #13m# with the help of #costheta # component of the initial velocity.
Therefore, time of flight #t# is given by the relation
#t=13/(ucostheta)# ......(1)
Landing angle of the ball indicates that the ball moves down as it enters the hoop. #x#-component of the velocity remains unchanged.
#:.11.45xxcos30.8^@=ucostheta#
#=>u cos theta=9.8# ......(2)
From (1) we have
#t~~1.32s#
Since #x and y# components of velocity are orthogonal, these can be dealt with independently.
Kinematic equation in the #y# direction is given
#v=u+at#
Inserting given values and taking #g=9.8ms^-2#, we get
#-11.45xxsin 30.8^@=usin theta-9.8xx1.32#
#-ve# sign in front of #g# indicates that it is opposing the direction of initial motion. Moreover, vertical component of velocity as the ball enters hoop is along the #-y# axis.
#=>-5.86=usin theta-13#
#=>usin theta=7.14# ......(3)
Dividing (3) by (2)
#tan theta=7.14/9.8#
#theta=36.1^@#, rounded to one decimal place
From (3)
#u=7.14/(sin36.1^@ )=12.1ms^-1#, rounded to one decimal place.
