# Question #8544f

Jul 31, 2016

The solution lists out steps and needs verification.

#### Explanation:

Let the basket ball be launched from origin with initial velocity of $\vec{u}$ which makes an angle $\theta$ with $x$-axis.
Also assuming negligible friction, the distance traveled in $x$ direction is $13 m$ with the help of $\cos \theta$ component of the initial velocity.

Therefore, time of flight $t$ is given by the relation
$t = \frac{13}{u \cos \theta}$ ......(1)

Landing angle of the ball indicates that the ball moves down as it enters the hoop. $x$-component of the velocity remains unchanged.
$\therefore 11.45 \times \cos {30.8}^{\circ} = u \cos \theta$
$\implies u \cos \theta = 9.8$ ......(2)

From (1) we have
$t \approx 1.32 s$

Since $x \mathmr{and} y$ components of velocity are orthogonal, these can be dealt with independently.
Kinematic equation in the $y$ direction is given

$v = u + a t$
Inserting given values and taking $g = 9.8 m {s}^{-} 2$, we get

$- 11.45 \times \sin {30.8}^{\circ} = u \sin \theta - 9.8 \times 1.32$
$- v e$ 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.

$\implies - 5.86 = u \sin \theta - 13$
$\implies u \sin \theta = 7.14$ ......(3)

Dividing (3) by (2)
$\tan \theta = \frac{7.14}{9.8}$
$\theta = {36.1}^{\circ}$, rounded to one decimal place

From (3)
$u = \frac{7.14}{\sin {36.1}^{\circ}} = 12.1 m {s}^{-} 1$, rounded to one decimal place.