Answer:
1.1 s
Explanation:
The initial ball’s height is [tex]3.5gsin(80.6)^{\circ}=3.453003g m\approx 3.45g m[/tex]
From the law of conservation of energy, the potential energy converts to kinetic energy
[tex]PE=mgh[/tex]
[tex]KE= 0.5mv^{2}[/tex]
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, [tex]I=\frac {2mr^{2}}{3} but [tex]w=\frac {v}{r}[/tex] and making r the subject then substituting it back to the equation of rotation energy we obtain rotational energy as [tex]\frac {mv^{2}}{3}[/tex]
[tex]3.5sin(80.6)^{\circ}=0.5mv^{2} + \frac {mv^{2}}{3}[/tex]
[tex]3.5sin(80.6)^{\circ}=\frac {5v^{2}}{6}[/tex]
[tex]v^{2}=\frac {21g sin(80.6)^{\circ}}{5}[/tex]
From kinematics
[tex]v^{2}=2as=2\times 3.5\times a[/tex]
[tex]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}[/tex]
Since [tex]s=0.5at^{2}[/tex] then [tex]t^{2}=\frac {2s}{a}[/tex] and [tex]t=\sqrt {\frac {2s}{a}}=\sqrt {\frac {2\times 3.5}{5.806964 m/s^{2}}}=1.09793 s\approx 1.1 s[/tex]
