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.
