Answer:
The speed of the ball is 5.59 m/s.
Explanation:
Given that,
Mass = 0.160 kg
Radius = 70 cm =0.70 m
Distance = 1.50
Speed at top = 3.26 m/s
Angle = 30.0°
We need to calculate the speed of the ball
The total energy at the top
[tex]K.E_{i}+P.E=\dfrac{1}{2}mv_{i}^2+mgh[/tex]
The final kinetic energy of ball at that point when the cord is 30° below the horizontal
[tex]K.E_{f}=\dfrac{1}{2}mv_{f}^{2}[/tex]
Using conservation of energy
[tex]K.E_{f}=K.E_{i}+P.E[/tex]
[tex]\dfrac{1}{2}mv_{f}^2=\dfrac{1}{2}mv_{i}^2+mgh[/tex]
[tex]\dfrac{1}{2}\times0.160\times v_{f}^2=\dfrac{1}{2}\times0.160\times(3.26)^2+0.160\times9.8\times(0.70+0.70\times\sin30^{\circ})[/tex]
[tex]v_{f}=\sqrt{(3.26)^2+2\times9.8\times1.05}[/tex]
[tex]v_{f}=5.59\ m/s[/tex]
Hence, The speed of the ball is 5.59 m/s.
