So I assume the angle of whirling is parallel to the ground.
Time to fall 2.8 m: 2.8=4.9t^2
solve for time in air.
29m=Vi*timeinair
solve for Vi.
then, centripetal acceleration
You plan to throw stones by using a sling of length 1.8 m which you whirl over your head. Suppose you wish to throw a stone a distance of 29 m. What must be the centripetal acceleration of the stone just before its release if it is to reach this distance? Assume that the release height is 2.8 m.
2 answers
thanks that makes sense