Physics
posted by Emily on .
A ball on the end of a string is whirled around
in a horizontal circle of radius 0.486 m. The
plane of the circle is 1.83 m above the ground.
The string breaks and the ball lands 2.39 m
away from the point on the ground directly
beneath the ball’s location when the string
breaks.
The acceleration of gravity is 9.8 m/s^2
Find the centripetal acceleration of the ball
during its circular motion.

a=mv²/R=?
h=gt²/2
t=sqrt(2h/g) = sqrt(2•1.83/9.8)=0.61 s.
L=v•t = >
v=L/t=2.39/0.61=3.92 m/s
a= mv²/R= m•(3.92)²/ 0.486 =…
You need mass ‘m’