Answer:
How far horizontally from point A does the ball hit the floor d= 5 m
Explanation:
By law of conservation of energy we can say that
final Kinetic energy kf+ final potential energy Uf= initial Kinetic energy ki+ initial potential energy Ui
We know that the ball has Uf, the ball at the bottom and no Ki ,the ball at initial state
therefore,
Kf= Ui
therefore,
[tex]\frac{1}{2}I\omega_f^2+\frac{1}{2}mv_f^2= mgh_i[/tex]
MOI for Solid ball = [tex]\frac{2}{5}mR^2[/tex]
[tex]\frac{1}{2}\frac{2}{5}mR^2\frac{V}{R}^2+\frac{1}{2}mV^2= mgh_i[/tex]
=[tex]\frac{7}{10}mV^2= mgh_i[/tex]
[tex]V= \sqrt{\frac{10}{7}gh_i}[/tex]
Now substituting the values to get value of V
[tex]V= \sqrt{\frac{10}{7}9.81\times4}[/tex]
V= 7.48 m/s
now to calculate How far horizontally from point A does the ball hit the floor
h= [tex]\frac{1}{2}gt^2[/tex]
2= [tex]\frac{1}{2}gt^2[/tex]
t=0.6385 sec
now horizontal distance d= vt = 0.6385×7.48= 5.00 m
