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.