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## Homework Statement

The string (in the pic) is L=120 cm long,has a ball attached to one end,and is fixed as its other end.The distance d from the fixed point end to a fixed peg at point P is 75.0cm.When the initally stationary ball is released with the string horizontal as shown,it will swing along the dashed arc.Whats its speed when it reaches (a) its lowest point and (b) its higest point after the string catches on the peg ?

## Homework Equations

##(ΔU)=U_f-U_i##

##-ΔU=W##

##W=\frac 1 2 m((v_f)^2-(v_i)^2)##

## The Attempt at a Solution

[/B] İts lowest point is in A.Using;

##-ΔU=W##

##W=\frac 1 2 m((v_f)^2-(v_i)^2)##

we get

##-mgH+mg(H+d+r)=\frac 1 2m(v_f)^2-0##

##2gL=(v_f)^2##

##2.9.8\frac {m} {s^2}.1.2m=(v_f)^2##

##v_f=4.85 \frac m s##

which its speed in the lowest point.And answer for part (a)

For part (b),The object has inital speed ##v_i=4.85 \frac m s## and goes point B.

same equations;

##-mg(H+r)+mg(H)=\frac 1 2 m((v_f)^2-(v_i)^2)##

##\frac 1 2m(v_i)^2-mgr=\frac 1 2 m(v_f)^2##

If we multiply by 2 and divide m we get

##(v_i)^2-2gr=(v_f)^2##

##((4.85 \frac m s)^2)-(2.9.8\frac {m} {s^2}.0.45m)=(v_f)^2##

##23.52\frac {m^2} {s^2}-8.82\frac {m^2} {s^2}=(v_f)^2##

##v_f=3.83\frac m s##

which book says ##v_f=2.42\frac m s##

Where did I go wrong ?

Thanks