A 88.0 kg boulder is rolling horizontally at the top of a vertical cliff that is 20 m above the surface of a lake, as shown in the figure . The top of the vertical face of a dam is located 100 m from the foot of the cliff, with the top of the dam level with the surface of the water in the lake. A level plain is 25m below the top of the dam. (height=h=45m, distance=deltaX=100m)
A)What must be the minimum speed of the rock just as it leaves the cliff so it will travel to the plain without striking the dam?
B)How far from the foot of the dam does the rock hit the plain?
(3) Yf=Yo + Vo,y(t) +.5*a*t^2
(4) Xf,x=Xf + Vo,x(t) + .5*a,x*t
The Attempt at a Solution
I had figured the origin would be at the top of the dam. using the eqn 3, with Yo= 0 and Vo,y=0, i could use that to solve for time: t=sqrt((2*h)/g)= 3.0secs. with time i could find the Vo,x because it states how far over the dam it must fly.
My original answer of part A was incorrect of 33.0 m/s and without it i have no clue how to set up B