Ball A is dropped from the top of a building of height h at the same instant that ball B is thrown vertically upward from the ground. At what height will the balls collide if the collision occurs when the balls are moving in the same direction and the speed of A is 4 times that of B.
VA0 = 0 - the internal speed of ball A.
tc - the time when the balls collide.
v = v(initial) +at
x = x(initial) + v(initial)t + (at^2)/2
The Attempt at a Solution
XA(tc)=XB(tc)=hc (height whenn the balls collide)
XB(tc)= XB(tb) - XB(tb→tc)=
=VBO - (g*(tb)2)/2 - (g*(tc-tb)2)/2=
=VBO - (g*(tb)2) - (g*(tc)2)/2 + g*tc*tb
⇒XB(tc)=g*tc*tb - (g*(tc)2)/2
VA(tc)=4*VB(tc) ⇒ -g*tc = -g*(tc-tb) ⇒ 3tc= 4tb →witch is impossible!