hit at height x
ball A drops h-x
ball b rises x
Ball A
(h-x) = h - 4.9 t^2 or x = 4.9 t^2
v = -9.81 t
Ball B
h = Vi t - 4.9 t^2
v = 9.81 t/2 = 4.9 t
so
4.9 t = Vi - 9.81 t
or Vi = 14.7 t
------------------
h = 14.7 t^2 - 4.9 t^2 = 9.8 t^2
but x = 4.9 t^2
so I get half way up, not 2/3
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. When the balls collide, they are moving in opposite directions, and the speed of A is twice the speed of B. At what height does the collision occur? (Use any variable or symbol stated above as necessary.)
I'm not sure if this is a question in which you need an equation or just a conceptual question, but I thought it would be 2/3, but webassign (our online hw assignments) keeps telling me that it is a mistake. Could you show me what I am doing wrong?
1 answer
1 answer