Let the basket ball be launched from origin with initial velocity of vecu→u which makes an angle thetaθ with xx-axis.
Also assuming negligible friction, the distance traveled in xx direction is 13m13m with the help of costheta cosθ component of the initial velocity.
Therefore, time of flight tt is given by the relation
t=13/(ucostheta)t=13ucosθ ......(1)
Landing angle of the ball indicates that the ball moves down as it enters the hoop. xx-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.
