Question

A regulation basketball has a 47 cm diameter and may be approximated as a thin spherical shell.

Answer 1

Answer:

1.1 s

Explanation:

The initial ball’s height is $3.5gsin(80.6)^{\circ}=3.453003g m\approx 3.45g m$

From the law of conservation of energy, the potential energy converts to kinetic energy

$PE=mgh$

$KE= 0.5mv^{2}$

Where m is mass, g is acceleration due to gravity, h is height, v is the velocity

Here, the potential energy converts to potential energy and rotation energy

Moment of inertia of the ball, $I=\frac {2mr^{2}}{3} but [tex]w=\frac {v}{r}$ and making r the subject then substituting it back to the equation of rotation energy we obtain rotational energy as $\frac {mv^{2}}{3}$

$3.5sin(80.6)^{\circ}=0.5mv^{2} + \frac {mv^{2}}{3}$

$3.5sin(80.6)^{\circ}=\frac {5v^{2}}{6}$

$v^{2}=\frac {21g sin(80.6)^{\circ}}{5}$

From kinematics

$v^{2}=2as=2\times 3.5\times a$

$a=\frac {v^{2}}{7}=\ frac {21g sin(80.6)^{\circ}}{5\times 7}=\ frac {21\times 9.81 sin(80.6)^{\circ}}{5\times 7}=5.806964 m/s^{2}$

Since $s=0.5at^{2}$ then $t^{2}=\frac {2s}{a}$ and $t=\sqrt {\frac {2s}{a}}=\sqrt {\frac {2\times 3.5}{5.806964 m/s^{2}}}=1.09793 s\approx 1.1 s$